DRAWING THE PSEUDO-ARC
WAYNE LEWIS AND PIOTR MINC
Abstract. It is very likely that the pseudo-arc may occur as an attractor
of some natural dynamical system. How would a picture of such a strange
attractor look? Would it be recognized as the pseudo-arc, a hereditarily indecomposable continuum? This paper shows that it could be difficult. We
notice that no black and white image can look hereditarily indecomposable on
any raster device (like a computer screen or a printed page). We also try to
generate the best computer picture of the pseudo-arc as it is possible under
the circumstances. With that purpose in mind, we expand the pseudo-arc into
an inverse limit with relatively simple, deterministically defined and easy to
handle numerically n-crooked bonding maps. We use this expansion to assess
numerical complexity of drawing the pseudo-arc with help from the AndersonChoquet embedding theorem. We also generate graphs of n-crooked maps
with large n’s, and we prove that a rasterized image of such a graph does not
look very crooked at all because it must contain a long straight linear vertical
segment.
1. Introduction
1.1. Motivation for this paper. The pseudo-arc was initially investigated by
Knaster [22], Moise [30], [31] and Bing [3],[4], [6]. It has been investigated extensively in subsequent years and has appeared in a number of research areas. There
are numerous significant questions concerning it still of interest and to be investigated.
The pseudo-arc is considered by many to be a very complicated object. This may
be in part due to the fact that it has structural properties quite different from those
of more familiar topological spaces. It is the simplest nondegenerate hereditarily
indecomposable continuum. Not only does it not contain any arcs, but it does not
contain any two non-nested subcontinua which intersect.
However, the pseudo-arc can be constructed by an easily described iterative
process satisfying a few general conditions. Much of the work on the pseudo-arc
has utilized a construction by a sequence of nested chain covers. It can also be
described in terms of an inverse limit of arcs. There is a direct correspondence
between the two descriptions since the nerves of the open covers given by the chains
are arcs.
In this presentation we use an inverse limit construction. This allows specificity
in the description of bonding maps and in computations. Inverse limits are probably
also more familiar to a larger audience.
In 1984, in response to questions about the ability to use computer graphics
to illustrate the pseudo-arc, the first author presented [26] brief observations on
the extremely rapid growth in the number of elements in crooked chains in just a
2000 Mathematics Subject Classification. Primary 54F15, 54-04;Secondary 54H20.
Key words and phrases. Pseudo-arc, pseudoarc, hereditarily indecomposable continuum, computer generated image, raster graphics, strange attractors, crooked map.
1
2
WAYNE LEWIS AND PIOTR MINC
few stages of the construction of the pseudo-arc, and observations on the severe
limitations on the ability to realistically illustrate this.
In 2004, the Canadian artist Dermot Wilson, with mathematical input from
Professor Murat Tuncali of the University of Nipissing, in North Bay, Ontario,
developed a multimedia presentation Pseudo-arc 4 the People. It was claimed,
without specific expansion on the claim, that part of the goal of the project was
to attempt “to visualize a plausible but undrawable object, pseudo-arc.” Most
of the artistic components of the project consisted of various representations of a
considerable simplification of crooked patterns.
In 2007, in a plenary address at the Third International Meeting on Continuum
Theory in Puebla, México, the second author presented several observations on
the complexity and extremely rapid growth in number of elements in inverse limit
constructions of the pseudo-arc and limitation in the ability to realistically draw
the pseudo-arc.
This paper expands on these observations.
1.2. Basic notions. Every space considered in this paper is metric and every map
is continuous. A continuum is a connected compact space. A chain is a collection
of open sets (called links) {C0 , C1 , . . . , Cn } such that Ci ∩ Cj 6= ∅ if and only if
|i − j| ≤ 1. A continuum X is chainable if, for each ² > 0, there is a chain covering
of X with links of diameter less than ².
Let I denote the unit interval [0, 1]. An arc is a continuum homeomorphic to I.
A continuum X is chainable if and only if it can be expressed as the inverse limit
of a sequence of of arcs (see [33, 12.11 and 12.19]).
1.3. A method of embedding chainable continua in the plane. We briefly
describe here a well-known method of embedding chainable continua in the plane.
The chain version of this method was first described by by Bing [6], and the inverse
limit approach (in more general settings) was given by Isbell [14]. We start with
the following proposition.
Proposition 1.1. Suppose ² is a positive number, f : I → I is a map, and h is an
embedding of I in the plane. Then, there is an embedding h0 of I in R2 such that
|h (f (t)) − h0 (t)| < ² for each t ∈ I.
¤
Consider the sequence I0 , I1 , I2 , . . . of copies of I. Suppose fi : Ii → Ii−1 is a
∞
map for each i = 1, 2, . . . . Denote by X the inverse limit lim (Ii−1 , fi )i=1 . If k < i,
←
−
let fki denote the composition fk+1 ◦ fk+2 ◦ · · · ◦ fi .
We construct a sequence of embeddings h0 , h1 , . . . of I into R2 by induction.
Let h0 (t) = (0, t) for each t ∈ I. Now, suppose hi−1 has been constructed for
some positive integer i. Pick any ²i > 0 and use Proposition 1.1 to construct an
embedding hi : I → R2 such that |hi−1 (fi (t)) − hi (t)| < ²i . Set gi = hi−1 ◦fi ◦hi −1
∞
and observe that lim (hi−1 (Ii−1 ) , gi )i=1 is homeomorphic to X. By choosing ²i to
←−
be sufficiently small (depending on the previous construction) one can satisfy the
hypotheses of the Anderson-Choquet Embedding Theorem [1, Theorem I] (see also
[33, 2.10]). It follows from the theorem that the limit (in the topological sense) of
the sequence h0 (I0 ) , h1 (I1 ) , . . . is homeomorphic to X. In practical terms, for a
sufficiently large integer i, the arc hi (Ii ) should be a good approximation of X.
Plotting hi (Ii ) should generate a good picture of X.
DRAWING THE PSEUDO-ARC
3
Let πy be the projection of R2 onto the y-axis. Observe
¯ that, for each δ¯ > 0,
the above construction can be done in such a way that ¯πy (hi (t)) − f0i (t)¯ < δ.
Therefore one can view the arc hi (Ii ) as a horizontal rearrangement of the graph
of f0i .
1.4. Hereditary indecomposability. A continuum X is indecomposable if it cannot be expressed as the union of two of its proper subcontinua. A continuum is
hereditarily indecomposable if each of its subcontinua is indecomposable. The first
nondegenerate chainable hereditarily indecomposable continuum was constructed
by Knaster in 1922 [22]. In 1948, Moise [30] constructed a nondegenerate chainable
hereditarily indecomposable continuum that was similar to an arc in the sense that
it was homeomorphic to each of its nondegenerate subcontinua. Moise named his
continuum a pseudo-arc. Also in 1948, Bing [3] proved that a construction similar to
that by Moise yields a homogeneous continuum. Next year, Moise [31] proved that
his original pseudo-arc was homogeneous. In 1951, Bing [4] proved that all nondegenerate chainable hereditarily indecomposable continua were homeomorphic. So
the pseudo-arc is topologically unique.
Hereditary indecomposability is equivalent to crookedness. Roughly speaking,
a continuum X is crooked if in order to get from an arbitrary point x ∈ X to
an arbitrary point y ∈ X, one first has to go near y, then return close to x before
finally reaching y (see [5], [7], [23] and [24]). In most constructions of the pseudo-arc,
hereditary indecomposability is achieved by introducing crookedness. To construct
the pseudo-arc by a sequence of nested chains C0 , C1 , C2 , . . . , it is usually assumed
that Ci is crooked in Ci−1 for each positive integer i (see [27] and [18]). To construct
the pseudo-arc by an inverse sequence, one makes the bonding maps fij ²j -crooked
with limj→∞ ²j = 0. A map f : I → I is ²-crooked if, for each a and b such that
0 ≤ a < b ≤ 1, there are c and d such that a < c < d < b, |f (b) − f (c)| < ² and
|f (a) − f (d)| < ² (see Definition 3.1 and Proposition 3.2).
In constructing the pseudo-arc as an inverse limit of arcs, it is not necessary that
each individual bonding map be crooked. In 1964, Henderson [11] constructed the
pseudo-arc as an inverse limit of copies of [0, 1] where every single-step bonding map
fii+1 was the same map f . The construction may be described roughly as starting
with g(x) = x2 and notching its graph with an infinite set of non-intersecting v’s
which accumulate at (1, 1) (see [11]). One by one, the notches are stretched by
iterations of f producing the required crookedness. The map f is C ∞ on [0, 1). By
essentially the same construction, one can obtain the pseudo-arc as an inverse limit
of copies of [0, 1] with one bonding map whose graph consists of countably many
straight linear segments.
One can also construct the pseudo-arc as an inverse limit with different bonding
maps which are unimodal or bimodal, introducing one or two folds at a time. The
idea behind such constructions is to use the composition of a very large number of
these simple maps to to get a single bonding map with a given degree of crookedness.
Folds between pairs of open sets are introduced whenever they are needed to make
the compositions more and more crooked.
Usually, constructions of the pseudo-arc are not explicit. To get the next chain
in the chain construction, or to get the next bonding map in the inverse limit
one simply requires that the next stage is sufficiently crooked in the preceding
stage. This lack of specificity makes any algorithmic approach to generating pictures
difficult.
4
WAYNE LEWIS AND PIOTR MINC
1.5. Raster graphics and relative decomposability number. In this paper we
investigate whether the pseudo-arc or any other plane continuum can be realistically
rendered on a computer screen or a printer. We assume that the screen (or the
printer) is a monochrome (black-and-white) raster device (see [9, p. 2], and sections
5 and 6 in this paper for more details), that is a rectangular array of intensity spots
called pixels. Each of the pixels can be individually addressed and set to either black
or white. We denote the collection of all pixels by D. For simplicity we assume that
the elements of D are congruent s by s squares tiling a rectangle R which is wd ·s wide
and hd · s high. This means that D = {Pij | 0 ≤ i ≤ wd − 1 and 0 ≤ j ≤ hd − 1}
where i and j describe the horizontal and the vertical, respectively, positions in
pixels of Pij in R, counting from the left lower corner of R. The numbers wd and
hd are called the horizontal and vertical resolution of the raster device. By the griddistance between two pixels Pi0 j 0 and Pi00 j 00 we understand the maximum of |i0 − i00 |
and |j 0 − j 00 | (see Definition 6.1). For any nonempty collection K ⊂ D, let d (K)
denote the maximal grid-distance between two pixels in K. By the grid-diameter of
K we understand d (K) + 1 (see Definition 6.1). Observe that the grid-diameter of
K is less than or equal to the minimum of wd and hd . We say that K is connected
if its union is connected (see Definition 6.1). If k ≥ 0 is an integer, then (K, k)
denotes the collection of those elements of D whose grid-distance to some element
of K is less than or equal to k (see Definition 6.1).
The rasterized (converted to pixels) image is the collection P ⊂ D of those pixels
that are set to black. Let n be the grid-diameter of P. Let k be the greatest integer
with the property that there are two connected collections A0 ⊂ P and A00 ⊂ P such
that A0 ∩ A00 6= ∅, A0 \ (A00 , k − 1) 6= ∅ and A00 \ (A0 , k − 1) 6= ∅. Equivalently, k is
the greatest integer such that the collection P contains two intersecting connected
clusters of pixels sticking out of each other (both horizontally and vertically) by at
least k · s. By the relative decomposability number of P we understand the ratio
k/n (see Definition 6.4).
The relative decomposability number is a measure of how decomposable connected subcollections of the rasterized image appear to be. If this number, considered as a fraction of the diameter of the entire image, is big enough to be well-visible,
the image does not look hereditarily indecomposable. We prove that each connected
collection P ⊂ D with the grid-diameter n has the relative decomposability number
greater than 1/ (1 + 2 log2 n) − 1/n (see Theorem 6.3 and Corollary 6.5). The sequence (1/ (1 + 2 log2 n) − 1/n) converges to 0 slowly. It follows that no black and
white image can look hereditarily indecomposable on a raster device (a screen or a
printer) with a reasonable resolution (see Observation 6.6). Increasing n and consequently the resolution of the image to make 1/ (1 + 2 log2 n) − 1/n small enough
is not productive due to limitations of the human eye.
1.6. Drawing the pseudo-arc. As we noticed above, no computer generated picture can look hereditarily indecomposable. Even so, it would be interesting to
generate a good image of the pseudo-arc and see what it actually looks like. A
satisfactory algorithm should be able to represent the image internally either by
a geometric description, or by a high resolution bit-map. We would then view a
rendering of this image through an imperfect display device ultimately restricted
by limitations of the human eye.
We investigate here the feasibility of writing such an algorithm by using the
method of embedding chainable continua in the plane described in subsection 1.3.
DRAWING THE PSEUDO-ARC
5
Before such an algorithm is written, one needs to express the pseudo-arc as an
inverse limit of copies of I with relatively simple and deterministically defined
bonding maps with growing degree of crookedness. For that purpose, we consider a
variation of ²-crookedness that is easier to handle numerically. We call this notion
n-crookedness where n is a positive integer. It differs from ²-crookedness described
above in that we take ² = 1/n and we require f (a) = i/n and f (b) = j/n for some
integers i and j (see Definition 3.3). We then construct the simplest n-crooked
surjection σn of I onto itself (Definition 3.8, see also Figures 2 and 3). σn is a
piece-wise linear map defined on cr [n] basic subintervals, each of which is taken by
σn onto an interval of the form [(i − 1) /n, i/n] for some positive integer i ≤ n. The
∞
crookedness sequence (cr [n])n=1 is a Fibonacci-like sequence defined by
• cr [1] = 1, cr [2] = 2 and
• cr [n] = 2cr [n − 1] + cr [n − 2] for each n ≥ 3.
We evaluate this sequence directly in Proposition 2.2.√In particular, it follows that
cr [n] is (almost) proportional to q n−1 where q = 1 + 2.
We consider a sequence of positive integers n1 , n2 , n3 , . . . such that ni+1 = 2cr[ni ]
for each positive integer i. By setting fi = σni , we get an expansion of the pseudoarc into an inverse limit with explicitly defined bonding maps (see Theorem 3.24).
We use this expansion to investigate the feasibility of drawing the pseudo-arc (see
the outline below). We also believe that the expansion may have meaningful applications in further theoretical studies of the pseudo-arc.
The sequence n1 , n2 , n3 , . . . is chosen rather conservatively. Any essentially
smaller growth would not produce the pseudo-arc. For instance, replacing the
condition ni+1 = 2cr[ni ] by ni+1 = cr[ni ] would result in the inverse limit containing arcs. This conservative choice notwithstanding, the sequence n1 , n2 , n3 , . . .
grows extremely rapidly. The numbers are more than astronomical, see Remark
3.25. For instance, if we start with n1 = 3, the map f01 is defined on cr [3] = 5
basic subintervals stretching each of them from the length of 1/5 to the length
of 1/3. Since n2 = 2cr [3] = 10, the map f12 = σ10 is defined on cr [10] = 2378
basic subintervals (see Figure 3) mapping each them to an interval of the form
[(i − 1) /10, i/10] where i = 1, . . . , 10. f01 = σ5 stretches this last interval to the
length of 1/6. Consequently, f02 linearly stretches each of the cr [10] basic intervals
for f12 = σ10 to the length of 1/6. Continuing this process one step further, we get
n3 = 2cr [10] = 4756. Using Proposition 2.2, one can estimate cr [4756] ≈ 101820 .
So, f23 = σ4756 maps each basic interval for f23 = σ4756 onto an interval of the form
[(i − 1) /4756, i/4756] where i = 1, . . . , 4756. Since 4756 = 2cr [10], f12 = σ10 maps
this last interval onto an interval of the form [(j − 1) /20, j/20] where j = 1, . . . , 20.
f01 = σ5 stretches this last interval to the length of 1/12. Consequently, f03 linearly
stretches each of the cr [4756] basic intervals for f23 = σ4756 to the length of 1/12.
Suppose, for the sake of argument, that the method described in subsection 1.3 is
used to generate a picture of the pseudo-arc by plotting the image of h3 . (This is
strictly an academic assumption in view of the magnitude of 101820 .) Suppose also
that ²1 , ²2 and ²3 are small in comparison with 1/12, and A is any of the intervals basic for f23 = σ4756 . Then h3 (A) is very close to a straight linear segment
whose projection on the y-axis is 1/12-long. Since 1/12 of the height of the picture
is well-visible, starting with n1 = 3 certainly does not lead to a good picture of
the pseudo-arc. Increasing n1 only slightly would not help by a similar argument.
One needs to start with n1 big enough so that 1/n1 is not perceived as a visible
6
WAYNE LEWIS AND PIOTR MINC
distance. But such n1 , in any reasonable resolution, would be in hundreds if not
in thousands, making n2 astronomically huge and completely preventing a plot of
the image of h2 . Thus, generating a realistic picture of the pseudo-arc by using the
method based on the the Anderson-Choquet theorem (subsection 1.3) appears to
be not feasible.
Observe that the graph of no map f : I → I looks like a hereditarily indecomposable continuum. For instance, the graph of f may be represented by the union
of two arcs F1 and F2 that intersect at one point and whose Hausdorff distance is
at least 0.5. (Set F1 and F2 to be the graphs of f restricted to [0, 0.5] and [0.5, 1],
respectively.) However, if g is an n-crooked map of I onto itself, and A and B are
intersecting arcs contained in the graph of g, then one of πy (A) or πy (B) must
be contained in the 1/n-neighborhood of the other. So, for sufficiently large n, the
graph of any n-crooked map should look hereditarily indecomposable with respect
to the y-axis. Certainly, one would not expect to see in an image of such graph
any long vertical segments. But, it turns out that, in this case, the reality is quite
different from the expectations. In sections 4 and 5 we examine how graphs of
n-crooked maps look on raster devices such as computer screens or printers. We
observe that, for large n, the rasterized black and white image of the graph of the
simple n-crooked map σn is either completely black or almost all black with two
symmetric irregularly shaped narrow white strips, one at the bottom and one at
the top (see Figure 5 (b) and Proposition 4.1). In general, we prove that if n is sufficiently large, then the image of the graph of any n-crooked map, when rasterized
in a reasonable resolution, must contain a long vertical segment (see Theorem 5.2
and Corollary 5.4).
Observe that if g is a map of I into itself which is not (n + 1)-crooked, then there
are two intersecting arcs A and B contained in the graph of g such that neither of
πy (A) nor πy (B) is contained in the 1/ (n + 1)-neighborhood of the other. So, if g
is n-crooked but not (n + 1)-crooked, then either n is small enough for 1/ (n + 1)
of the height of the graph to be visible, or n is large. In the former case, the
graph of g does not appear to be hereditarily indecomposable with respect to the
y-axis. However, Theorem 5.2 implies that, even for moderate choices of n, the
rasterized graph of g contains long vertical segments, and again does not appear to
be hereditarily indecomposable.
We include the two following examples. In Example 5.6 we show an n-crooked
map whose graph rasterized under many magnifications consists of one horizontal
row of pixels on the bottom and one vertical column on the right. In Example 5.7,
we show a n-crooked function g whose rasterized graph, restricted to some horizontal strip S = I × [a, 1 − a], looks like the ternary Cantor set cross [a, 1 − a]. When
the horizontal resolution is slowly increased, the rasterized graph of g restricted to
S is exactly the same as the rasterization the Cartesian product of the Cantor set
and [a, 1 − a].
1.7. Pseudo-arc and attractors of maps of the plane. In 1976, Hénon [12]
gave an example of a homeomorphism H of the plane onto itself such that orbits of
points from a certain region R converged to a strange continuum. One may view
the continuum by taking a point z ∈ R and plotting its forward orbit starting from
a certain iteration H n (z), see Figure 1(a). Parts of the picture can be magnified
by plotting only the iterations contained in a certain rectangle. Figure 1(b) shows
DRAWING THE PSEUDO-ARC
(a)
(b)
7
(c)
Figure 1. Hénon Attractor
the rectangle to be magnified with the magnification shown in Figure 1(c). Can
one generate an image of the pseudo-arc in a similar way?
In 1990, Barge and Martin [2] proved that for each continuous surjection f :
I → I there is a homeomorphism Hf of the plane onto itself with a global attractor
homeomorphic to the inverse limit lim (I, f ) of copies of I with all bonding maps
←−
equal to f . The dynamics of Hf restricted to A are analogous to those of the map
fˆ on lim (I, f ) induced by f . As we mentioned earlier, Henderson [11] constructed
←−
a map f : I → I such that lim (I, f ) is the pseudo-arc. Henderson’s map has two
←−
fixed points 0 and 1. The forward orbit of any t ∈ [0, 1) is attracted to 0. It
follows that, for every x ∈ lim (I, f ) different from (1, 1, 1, . . . ), the forward orbit of
←−
fˆ converges to (0, 0, 0, . . . ). Consequently, most forward orbits of Hf converge to a
single attracting point. Thus, the picture generated by iterating Hf , with f being
the Henderson map, would produce a single point instead of the pseudo-arc.
In 1991, Minc and Transue [29] constructed a transitive map g : I → I whose inverse limit is the pseudo-arc. For this map, most forward orbits of the Barge-Martin
homeomorphism Hg approximate the pseudo-arc. But, since g is complicated, it
would difficult to actually generate an image of the pseudo-arc based on this technique.
In 1951, Bing [4] defined a hereditarily indecomposable continuum separating
the plane. This continuum, called the pseudo-circle, has the property that each of
its proper nondegenerate subcontinua is a copy of the pseudo-arc. Therefore, an
image of the pseudo-circle would automatically yield an image of the pseudo-arc.
The pseudo-circle appears frequently in dynamical systems, see [10], [13], [19], [20],
[21], [28], [34] and [35]. Even though it is not apparent how to construct a map of
the plane suitable for generating a picture of the pseudo-arc or the pseudo-circle by
plotting iterations, it is very likely that each of these two hereditarily indecomposable continua may be an attractor for some dynamical system of the plane naturally
arising from applications of mathematics in some other field. Assume for the sake
of discussion that H is a map ofTthe plane with an attractor A homeomorphic to the
∞
pseudo-arc. Assume also that n=1 {H n (z) , H n+1 (z) , . . . } = A for some z ∈ R2 .
Finally, suppose that we have an algorithm to evaluate any iteration H i (z), and we
may view A or its parts as easily as we may view the Hénon attractor. How would
the picture of A then look? As we observed earlier, it would not look hereditarily
indecomposable, see Theorem 6.3, Corollary 6.5 and Observation 6.6. Would it be
possible in such a case to recognize the attractor as the pseudo-arc by looking at
8
WAYNE LEWIS AND PIOTR MINC
its plotted picture? We cannot answer this question definitely, but our experience
with drawing graphs of n-crooked maps gives us some indication of difficulties one
may encounter. First of all, it is very likely that we may see only a partial picture
of A, even though the forward orbit of z converges to the entire A. Simply, the
orbit of z may visit some parts of A less frequently than others. A similar situation
occurs when we attempt to draw the graph of the simple 200-crooked map σ200
by randomly choosing x ∈ I and then plotting (x, σ200 (x)), see section 4. That
“probabilistic” approach results in the plotted points being distributed in a narrow
horizontal band in the center of the unit square (see Figure 4), giving only a partial
picture of the graph (σ200 is a surjective map of I onto itself). But even if we could
see the full picture, would it be as distinctive as it was expected? Our experience
with drawing graphs of n-crooked maps suggests that the pseudo-arc, viewed in its
entirety and then under magnifications, may look either like a plane region (see
Figure 5 (b) and Proposition 4.1), or a single arc (see Example 5.6), or a bundle of
arcs (see Example 5.7).
2. The crookedness sequence
∞
In this section we introduce a Fibonacci-like sequence of positive integers (cr [n])n=1 .
We call this sequence the crookedness sequence and use it later in the paper to define
n-crooked maps and measure the degree of their complexity.
∞
Definition 2.1. Let (cr [n])n=1 be the sequence defined in the following way:
• cr [1] = 1, cr [2] = 2 and
• cr [n] = 2cr [n − 1] + cr [n − 2] for each n ≥ 3.
The crookedness sequence starts from 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741,
. . . and grows rapidly. For instance, one can calculate that cr [200] is 1269381235399
4620481037986488739368440399451028645237163046012909971924256728. The following proposition allows one to compute the crookedness sequence directly.
Proposition 2.2.
cr [n] = q n−1
(1)
where q = 1 +
³
√
1−
1
−1
q2
+ q12
´n
2.
Proof. Set r = −1/q 2 and observe that
(2)
2/q = 1 + r.
Observe also (1) is equivalent to
(3)
cr [n] (1 − r) = q n−1 (1 − rn ) .
Prove (3) by induction. For n = 1, (3) is obvious. For n = 2, (3) follows from (2).
Now, assume that (3) is true for each n less than a certain integer k ≥ 3, and prove
(3) for n = k.
¡
¢
¡
¢
cr [k] (1 − r) = (2cr [k − 1] + cr [k − 2]) (1 − r) = 2q k−2 1 − rk−1 +q k−3 1 − rk−2
Now, replace 2q k−2 by q k−1 (1 + r) (see (2)) and q k−3 by −q k−1 r to get
¡
¢
¡
¢
¡
¢
cr [k] (1 − r) = q k−1 (1 + r) 1 − rk−1 − q k−1 r 1 − rk−2 = q k−1 1 − rk .
The last equality is (3) for n = k.
¤
DRAWING THE PSEUDO-ARC
9
³
¢´−1
n+1 ¡
1 + 1/q 2
By neglecting (−q)
in (1), we get the following approximation:
(4)
cr [n] ≈
q n−1
1 + q12
In particular, it follows that cr[200] ≈ 1.26938 × 1076 .
3. n-crooked maps
In this section we introduce the notion of n-crookedness, define the simple ncrooked map σn and study its properties. As we promised in the introduction, the
simple n-crooked maps are deterministically defined and, therefore, easier to use in
numerical applications. In particular, numeric complexity of the simple n-crooked
maps are exactly described by the crookedness sequence defined in the previous
section. In Theorem 3.24, we explicitly define a sequence n1 , n2 , . . . and expand
the pseudo-arc into an inverse limit with the σni bonding maps.
Definition 3.1 (²-crookedness). [5], [7] Suppose that ² is a positive real number,
and g is a map of I onto itself. We say that g is ²-crooked if, for each a and b such
that 0 ≤ a < b ≤ 1, there are c and d such that a < c < d < b, |g (b) − g (c)| < ²
and |g (a) − g (d)| < ².
Proposition 3.2. [3], [4] Suppose
a nondegenerate continuum X is the inverse
¡
¢that
∞
limit of an inverse sequence Ik , gkl k=1 , where Ik = I for k = 1, 2, . . . and, for each
integers k and l such that l > k ≥ 1, gkl is a continuous surjection of Il onto Ik .
Then, X is a pseudo-arc if and only if for every ² > 0 and every positive integer k
there is an integer l > k such that gkl is ²-crooked.
It is convenient for us to divide I into n congruent strips and define crookedness
with respect to those strips. We call it n-crookedness. Even though the notation
for n-crookedness and ²-crookedness is similar, it will be always clear which of the
notions we mean. Simply, n will be always a positive integer while ² will be a
positive real number less than 1.
Definition 3.3 (n-crookedness). Let n > 0 be an integer, J ⊂ I be an interval, and
let g be a map of J into I. For each i, j ∈ {0, . . . , n} such that¡i <
¢ j, we say that
¡ g¢
is n-crooked between ni and nj provided that for each a ∈ g −1 ni and b ∈ g −1 nj
¡
¢
¡
¢
there exist a0 ∈ g −1 j−1
and b0 ∈ g −1 i+1
such that either a ≤ a0 ≤ b0 ≤ b or
n
n
0
0
a ≥ a ≥ b ≥ b. Observe that if i < j ≤ i + 2, then any g is n-crooked between i/n
and j/n.
We say that g is n-crooked if it is n-crooked between i/n and j/n for each
i, j ∈ {0, . . . , n} such that i < j.
The following two propositions establish the connection between ²-crookedness
and n-crookedness.
Proposition 3.4. Suppose g is a map of I onto itself. Let n be a positive integer
such that g is (1/n)-crooked in the sense of Definition 3.1. Then, g is n-crooked in
the sense of Definition 3.3.
¤
10
WAYNE LEWIS AND PIOTR MINC
σ5
(cr [5] = 29)
s5
σ6
1 − s5
(cr [6] = 70)
s6
1 − s6
Figure 2
Proposition 3.5. Suppose g is a map of I onto itself. Let ² > 0 and let n be a
positive integer such that 2/n < ² and g is n-crooked in the sense of Definition 3.3.
Then, g is ²-crooked in the sense of Definition 3.1.
¤
The following proposition is a simple consequence of Propositions 3.2, 3.4 and
3.5.
Proposition 3.6. Suppose
that
¡
¢∞ a nondegenerate continuum X is the inverse limit
of an inverse sequence Ik , gkl k=1 , where Ik = I for k = 1, 2, . . . and gkl : Il → Ik
for l > k ≥ 1. Then, X is a pseudo-arc if and only if for every pair of positive
integers n and k there is an integer l > k such that gkl is n-crooked.
¤
Definition 3.7. Let g1 and g2 be two maps of I into itself such that g1 (0) =
g2 (0) = 0 and g1 (1) = g2 (1) = 1. Suppose m ≥ 3 is an integer and s is a number
such that 0 < s < 0.5. Then ϕ [g1 , g2 , s, m] is the function of I into itself defined
by the formula:
 m−1 ¡ t ¢

 m g2 s ³
´ if 0 ≤ t ≤ s,
1
m−2
1−s−t
ϕ [g1 , g2 , s, m] (t) =
if s ≤ t ≤ 1 − s,
m + m g1 ¡ 1−2s ¢

 1
m−1
t+s−1
+
g
if
1 − s ≤ t ≤ 1.
2
m
m
s
The following definition is illustrated by Figure 2.
Definition 3.8 (simple n-crooked map σn ). Let
sn = cr [n − 1] /(2cr [n − 1] + cr [n − 2])
for each integer n ≥ 3. For each positive integer n, let the simple n-crooked map
σn be the map of I onto itself defined in the following way:
• σ1 = σ2 is the identity on I, and
• σn = ϕ [σn−2 , σn−1 , sn , n] for each integer n ≥ 3.
Let σ−n denote the reflection of the simple n-crooked map. (σ−n : I → I is defined
by σ−n (t) = 1 − σn (t).)
Proposition 3.9. σ−n (t) = σn (1 − t) for each t ∈ I.
¤
DRAWING THE PSEUDO-ARC
11
Proposition 3.10. The simple n-crooked map σn and its reflection σ−n are ncrooked.
¤
To simplify the notation, we set cr [−n] = cr [n] for any positive integer n.
Proposition 3.11. Let n be a nonzero integer, and let j be an integer such that 0 ≤
j < cr [n]. Then, there is an integer i such that 0 ≤ i < n and σn restricted to the
interval [j/cr [n] , (j + 1) /cr [n]] is a linear map onto the interval [i/n, (i + 1) /n].
¤
Intervals of the type [j/cr [n] , (j + 1) /cr [n]] are called the basic intervals for σn
and σ−n .
Proposition 3.12. Let n be a positive integer, let i 6= j be integers, and let [a, b] ⊂ I
such that
• σn (a) = i/n, σn (b) = j/n,
• one of i/n and j/n is the absolute minimum value of σn restricted to [a, b]
and the other is the absolute maximum value, and
• σn ((a, b)) ∩ {i/n, j/n} = ∅.
Then, σn restricted to [a, b] is a (rescaled and translated) copy of σj−i .
¤
Proposition 3.13. Let n be a positive integer, let i ∈ 1, . . . , n − 1, and let [a, b] be
a nondegenerate subinterval of I such that
• σn (a) = σn (b) = i/n, and
• σn ((a, b)) ∩ {i/n} = ∅.
Let m denote the midpoint of [a, b]. Then, σn −1 (σn (m)) ∩ [a, b] = {m}. (Observe,
that it follows that {i/n, σn (m)} is the set of the absolute maximum and minimum
values of σn restricted to [a, b].)
¤
Proposition 3.14. Let n be a positive integer. Let A = [a, b] × {t} be a horizontal
segment contained in the unit square below the graph of σn , and let i ≥ 0 be an
integer such that i/n ≤ t. Then, b − a ≤ cr [n − i] /cr [n].
Proof. Let [c, d] be the maximal subinterval of I such that [a, b] ⊂ [c, d] and (c, d) ×
{i/n} does not intersect the graph of σn . Observe that σn (c) = i/n and σn (d) =
1 or σn (d) = i/n. If σn (d) = 1, then Proposition 3.12 implies that d − c =
cr [n − i] /cr [n], and the proposition is true. So, we may assume that σn (d) = i/n.
Let m denote the midpoint of the interval [c, d]. By Proposition 3.13, σn restricted
to [c, d] has the unique absolute maximum at m. Let j be the integer such that
σn (m) = j/n. Observe that j < n. Using Proposition 3.12 twice, first on [c, m] and
then on [m, d], we get the result that d − c = 2cr [j − i] /cr [n]. Now, the proposition
follows, since 2cr [j − i] ≤ 2cr [n − 1 − i] < cr [n − i].
¤
Both σn and σ−n are the simplest continuous n−crooked surjections of I. We
introduce the following definition to measure complexity of arbitrary n−crooked
maps.
Definition 3.15 (scn ). Suppose g is a map of an arbitrary space into I. Let
n, i and j be integers such that 0 ≤ i < j ≤ n. We denote by scn [i, j, g] the
number of components of g −1 ((i/n, j/n)) that are mapped by g onto (i/n, j/n).
(The abbreviation sc stands for surjective components.) To simplify the notation,
we use scn [i, j] for scn [i, j, σn ], and scin for scn [i, i + 1, σn ].
12
WAYNE LEWIS AND PIOTR MINC
σ8
(cr [8] = 408)
σ10
(cr [10] = 2378)
Figure 3
Proposition 3.16.
Pn
k=0
sckn = cr [n] for each positive integer n.
¤
Proposition 3.17. Let n, i and j be integers such that 0 ≤ i < j ≤ n. Then,
scn [i, j] = scn [n − j, n − i] = scn [i, j, σ−n ] = scn [n − j, n − i, σ−n ].
¤
Proposition 3.18. Let n, i, j, k and l be integers such that 0 ≤ k ≤ i <
j ≤ l ≤ n. Suppose J ⊂ I is an interval minimal with respect to the property
σn (J) = [k/n, l/n]. Let g denote the restriction of σn to J. Then, scn [i, j, g] =
scl−k [i − k, j − k].
¤
Proposition 3.19. Let n, i, j, k and l be integers such that 0 ≤ k ≤ i < j ≤
l ≤ n. Suppose J ⊂ I is an interval and g : J → I is an n-crooked map such that
k l
¤
n , n ∈ g (J). Then, scn [i, j, g] ≥ scl−k [i − k, j − k].
Corollary 3.20. Let n, i and j be integers such that 0 ≤ i < j ≤ n. Suppose that
g is an n-crooked map of I onto itself. Then, scn [i, j, g] ≥ scn [i, j].
¤
Remark 3.21. Observe that the values of scin can be evaluated recursively using
the following equalities:
• sc0n = scnn−1 = 1 for n = 1, 2, . . . , and
i−1
• scin = scin−1 + scn−2
+ sci−1
¤
n−1 for n = 3, 4, . . . and i = 1, . . . , n − 2.
Proposition 3.22. Let n, k, l and µ be integers such that 0 ≤ k ≤ k + µ < l − µ ≤
l ≤ n. Suppose J ⊂ I is an interval and g : J → I is an n-crooked map such that
k l
µ
n , n ∈ g (J). Then scn [k + µ, l − µ, g] ≥ 3 .
Proof. We prove the proposition by induction with respect to µ. The proposition is
obvious for µ = 0. We assume that the proposition is true for some µ = i and prove
it for µ+1. For that purpose, we³may assume
´ that k +µ+1 < l −µ−1. ³Let (a, b) be
´
l−µ
l−µ
that is mapped by g onto k+µ
.
,
an arbitrary component of g −1 k+µ
n
n
n , n
To complete the proof
we prove that (a, b) contains
³ of the proposition,
´
³ at least three
´
k+µ+1 l−µ−1
−1 k+µ+1 l−µ−1
that
are
mapped
by
g
onto
.
components of g
,
,
n
n
n
n
We may assume without loss of generality that g (a) =
k+µ
n
and g (b) =
l−µ
n .
Since
DRAWING THE PSEUDO-ARC
13
g is n-crooked, there are a0 and b such that a < a0 < b0 < b, g (a0 ) = l−µ−1
and
n
0
0 0
0
g (b0 ) = k+µ+1
.
Observe
that
each
of
the
intervals
(a,
a
),
(a
,
b
)
and
(b
,
b)
contains
n
³
´
a subinterval mapped by g onto
k+µ+1 l−µ−1
, n
n
.
¤
Proposition 3.23. Suppose that n and k are positive integers. Let g be a continuous self-surjection of I which is (k · cr[n])-crooked. Then σn ◦ g and σ−n ◦ g are
(k · n)-crooked.
Proof. Let f be either σn or σ−n . Take integers i and j such that 0 ≤ i < i + 2 <
j ≤ kn, and prove that f ◦ g is (k · n)-crooked between i/ (kn) and j/ (kn). Let
a, b ∈ I be such that f ◦ g (a) = i/ (kn) and f ◦ g (b) = j/ (kn). To simplify the
notation we assume that a < b. The proof in the case a > b is essentially the same
and will be omitted. Let a0 be the greatest number in the interval [a, b] such that
f ◦ g (a0 ) = i/ (kn). Let b0 be the least number in the interval [a0 , b] such that
f ◦ g (b0 ) = j/ (kn). Let c and d denote g (a0 ) and g (b0 ), respectively. Clearly,
f (c) = i/ (kn), f (d) = i/ (kn) and f ([c, d]) = [i/ (kn) , j/ (kn)]. By Proposition
3.11, there are integers i0 , j 0 ∈ {0, 1, . . . , k · cr [n]} such that c = i0 / (k · cr [n]) and
d = j 0 / (k · cr [n]). We will make here another simplification and consider only the
case where i0 < j 0 . The proof in the remaining case is essentially the same and
will be omitted. Since i + 2 < j, it follows from Proposition 3.11 that i0 + 2 < j 0 .
Since g is (k · cr [n])-crooked, there exist a0 and b0 such that a0 < a0 < b0 < b0 ,
g (a0 ) = (j 0 − 1) / (k · cr [n]) and g (b0 ) = (i0 + 1) / (k · cr [n]). Since f ([c, d]) =
[i/ (kn) , j/ (kn)], it follows from Proposition 3.11 that f ◦ g (a0 ) = (j − 1) / (kn)
and f ◦ g (b0 ) = (i + 1) / (kn). This completes the proof of the proposition.
¤
Theorem 3.24. Let n1 , n2 , n3 , . . . be a sequence of positive integers such that
ni+1 = 2cr[ni ] for each positive integer i. Consider the sequence I0 , I1 , I2 , . . . of
copies of I. For each i = 1, 2, . . . , let fi : Ii → Ii−1 be either the simple ni ∞
crooked map σni or its reflection σ−ni . Then, the inverse limit of (Ii−1 , fi )i=1 is a
pseudo-arc.
Proof. For each pair of integers i and j such that 0 ≤ i < j, let fij denote the com¡
¢
position fi+1 ◦fi+2 ◦· · ·◦fj . It follows from Proposition 3.23 that fij is 2j−i−1 ni+1 crooked for each pair of integers i and j such that 0 ≤ i < j. Now, the theorem
follows from Proposition 3.6.
¤
Remark 3.25. The sequence n1 , n2 , n3 , . . . described in Theorem 3.24 grows rapidly.
For instance, if n1 = 2 (which makes f1 to be the identity on I), then n2 = 2cr [2] =
4, n3 = 2cr [4] = 24, n4 = 2cr [24] ≈ 9.0 × 109 and n5 = 2cr [n4 ] ≈ 1.7 × 10415954467 .
If we start with n1 = 3, we get n2 = 2cr [3] = 10, n3 = 2cr [10] = 4756 and
n4 = 2cr [4756] ≈ 2.1 × 101820 . Even that extra-astronomical growth is barely
enough to produce the pseudo-arc. Arcs are disappearing very slowly. Observe
that f0j restricted to each interval of ¡the form
¢ [k/cr [nj ] , (k + 1) /cr [nj ]] is a linear
map onto an interval with length 1/ 2j−1 n1 . For example, if n1 = 3, then f03 is a
piece-wise linear map defined on over 101820 intervals, stretching each of them from
the length less than 10−1820 to the length of 1/12 (see page 5 for more details).
It has been estimated that the number of elementary particles in the entire
observable universe is of the order 1079 , see [8, page 397]. The number is obtainable
from estimates of the average density of the observable universe, the size of the
observable universe and the mass of elementary particles. Allowing for uncertainty
14
WAYNE LEWIS AND PIOTR MINC
2000 points
5000 points
Figure 4. “Probabilistic” graphs of σ200
in the average density of the universe and for possibilities like dark matter, changes
this by at most one or two orders of magnitude. While these astronomical numbers
are very large, they are truly minuscule compared to the numbers obtained above
in just a few steps of forming crooked maps.
Observe also that changing the condition ni+1 = 2cr[ni ] in the statement of
Theorem 3.24 to ni+1 = cr[ni ] implies that each bonding map f0l is n1 −crooked and
one never gets greater crookedness to this first factor space by composing additional
bonding maps. Thus, such a sequence produces an inverse limit containing arcs,
and, therefore, not homeomorphic to the pseudo-arc.
4. Drawing the graph of the simple crooked map σ200
As we outlined in the introduction (see subsection 1.6) it appears to be not
feasible to generate a realistic picture of the pseudo-arc by using the method based
on the the Anderson-Choquet theorem described in subsection 1.3. Under these
circumstances, drawing the graph of σn , for sufficiently large n, may give us at
least some approximation of the pseudo-arc’s image.
The graph of σn consists of cr [n] basic straight linear segments, each of which
projects onto a (1/n)-long interval contained in the y-axis. Those straight linear
segments are very pronounced in Figures 2 and 3 depicting the graphs of σn for
n = 5, 6, 8 and 10. In the scale of the figures, the basic segments in the graph of σ10
are about 4 millimeters long. Since many of the neighboring segments are collinear,
we see a lot of straight linear segments that are 8 millimeters long. To defeat that
piece-wise linear appearance of the graph and get a glimpse into a picture of true
crookedness, we need to decrease the length of straight linear segments by scale of
at least 20. That would increase n to 200. In this section, we draw the graph of
σ200 . We start with a naive probabilistic approach (Figure 4). We then improve
our method to get a more interesting picture (Figure 5 (a)). Finally, we arrive at
our final picture of the graph of σ200 (Figure 5 (b)). The last picture is somewhat
disappointing, but it does reflect the limitations of perceiving the original abstract
mathematical object through human eyes.
DRAWING THE PSEUDO-ARC
(a)
15
(b)
Figure 5. Two more “graphs” of σ200
The graphs in Figures 2 and 3 are drawn as piecewise linear curves. The value
σn (xi ) is evaluated for each point xi = i/cr [n] where i = 0, 1, . . . , cr [n]. The consecutive points (xi , σn (xi )) are then joined by cr [n] segments. For example, the graph
of σ10 consists of cr [10] = 2378 basic segments. Since cr [n] grows exponentially
with n (see Proposition 2.2), the same approach cannot be used for higher values
of n. Certainly, not for n = 200 with cr [200] greater than 1.2 × 1076 . However,
even though the map σn is exponentially complicated, its value at any given point
may be computed in time proportional to n log n. This allows us to try the following probabilistic approach to drawing the graph of σ200 . We may choose a random
point x ∈ [0, 1], evaluate σ200 (x) and then plot the point (x, σ200 (x)). We may then
repeat this operation for many randomly chosen points x, and get an approximation
of the graph of σ200 . Figure 4 shows two such approximations, one consisting only
of 2000 plotted points, and the other containing an additional 3000 points. Observe
that there is not much difference in the appearance of those approximations. In
both of them, the plotted points are are distributed in a narrow horizontal band
in the center of the unit square. This is the result of probability. Indeed, one can
use the recursive formula in Remark 3.21 to compute, for each value a, the measure m (a) of I \ σ200 −1 ([0.5 − a, 0.5 + a]). For instance, m (0.1) ≈ 0.7 × 10−3 and
m (0.25) ≈ 1.4×10−17 . This last number makes it virtually impossible for the value
of σ200 at a randomly chosen point to be outside the interval [0.25, 0.75]. In order
to overcome this limitation, for a randomly chosen point x ∈ I, we compute x1 , the
first point at which σ200 has the absolute maximum on the interval [0, x]. Next, we
compute x2 , the first point at which σ200 has the absolute minimum on the interval
[x1 , x]. Then, we look for x3 , the first point at which σ200 has the absolute maximum
on the interval [x2 , x]. We continue this process until it ends with xix = x. We then
plot the points (x1 , σ200 (x1 )) , (x2 , σ200 (x2 )) , . . . , (xix , σ200 (xix )), and continue the
process with a new randomly chosen point x. Figure 5 (a) is a result of this procedure with 20000 plotted points based on a little over than 100 random choices of
x. The picture is intriguing, but still is not a satisfactory reflection of the graph of
σ200 . If we continued the procedure plotting millions of points (which would not be
a reasonable way of drawing a picture) we would get a picture similar to the one in
Figure 5 (b), almost all black with two symmetric irregularly shaped narrow white
16
WAYNE LEWIS AND PIOTR MINC
strips, one at the bottom and one at the top. To explain why the graph of σ200
looks as we claim, we need the following proposition.
Proposition 4.1. Suppose that i and n are integers such that 0 ≤ i < n. Let H
be a horizontal segment in I 2 below the graph of σn and
√
¡ on or¢ above the segment
I ×{i/n}. Then, the length of H is less or equal to q 2−i / q 2 − 1 , where q = 1+ 2.
Proof. It follows from Proposition 3.14 that the length of H is less than or equal to
cr [n − i] /cr
[n].
Applying Proposition 2.2 we get that the last number is equal to
³
´n−i
³ ´n−i
³ ´n
³ ´2
1− −1
2
³q ´n . Observe that 1− −1
z = q −i
≤ 1+ q12 and 1− −1
≥ 1− q12 .
−1
q2
q2
1−
q2
Using the last two inequalities we get that z ≤ q −i
1+ q12
³ ´2
1− q12
¡
¢
= q 2−i / q 2 − 1 .
¤
Suppose that every point of the graph of σn , for a certain positive integer n, has
been plotted and is visible as a black dot (disk) which is small but still big enough
to be discernable. Let D be a disk contained in the unit square outside of the
plotted graph. Since the graph is symmetric, we may assume that D lies below the
graph. Let H be the horizontal diameter of D, and let i be the integer such that
H ⊂ I × [i/n, (i + 1) /n). Now, Proposition 4.1 gives a correlation between i and
the length of H. The higher is i, the smaller, exponentially smaller, H must be.
For instance, if i = 10, the length of H and the diameter of D must be less than
0.00018. In the scale as in Figure 5 (b), where one unit corresponds approximately
to 40 mm, a white disk D, with its center 2 mm above the bottom of the unit
square, would have the diameter less than 0.0072 mm and would not be visible. In
the same picture, a disk centered only 1 mm above the bottom could have diameter
as much as 0.58 mm. Many such discs are visible in the figure.
Choosing n essentially smaller than 200 makes the graph of σn to be a poor
approximation of the pseudo-arc because the basic straight linear segments are visible. The choice of n = 200 (or similar) produces a picture with a lot of black space
and not much illustration of crookedness. If we increase n, the picture becomes
completely black.
5. n-crooked maps and raster graphics
In the previous section we observed that if the graph of σn (for large n) is displayed on a computer screen, the plotted points cover almost the entire screen. In
particular, the picture must contain long straight linear vertical segments. In this
section we extend the last observation to arbitrary n-crooked maps. To be able
to state and then prove our theorem in mathematical terms, we need to precisely
define how graphics are displayed. We assume that the screen (or the printer) is a
monochrome (black-and-white) raster device (see [9, p. 2]), that is a rectangular array of intensity spots called pixels. Each of the pixels can be individually addressed
and set to either black or white. Figure 6 illustrates how the graphs of σ5 , σ10 and
σ20 are displayed on a 20 by 20 “raster device.” The “actual” graphs of σ5 and
σ6 are also shown. The rasterized (converted to pixels) graphs of σ10 and σ20 do
not carry very much information, which is not surprising, since our raster device is
very crude. It has only 400 pixels, while the actual graphs of σ10 and σ20 consist of
cr [10] = 2378 and cr [20] = 15994428 basic segments, respectively. Actual display
and printing raster devices have much higher resolution. For example, a typical
DRAWING THE PSEUDO-ARC
17
σ5
σ6
σ10
σ20
Figure 6. Rasterized graphs of σ5 , σ6 , σ10 and σ20
modern 24” LCD desktop monitor has resolution of 1920 × 1200. A typical laser
printer can place 600 or 1200 dots of toner within an inch.
For the sake of simplicity, we assume that our display device is tiled by a wd by
hd grid of congruent square pixels Pij where i = 0, . . . , wd − 1 and j = 0, . . . , hd − 1.
Let s denote the length of each of the sides of Pij . Then Pij is the square in the
xy-plane with the left lower corner at (is, js) and its sides parallel/perpendicular
to the coordinate axes. Let pij be the intensity value of the pixel Pij , that is, in
our case, pij = 0 if Pij is white, and pij = 1 if Pij is black. In this context any
array of 0’s and 1’s is called a bit-map (see [9, p. 127]).
Definition 5.1. Suppose G ⊂ I 2 . Let w and h be positive integers. For each
i = 0, . . . , w − 1 and each j = 0, . . . , h − 1, let
Rij = {(x, y) | i/w ≤ x ≤ (i + 1) /w and j/h ≤ y ≤ (j + 1) /h} .
Set rij = 0 if Rij ∩ G = ∅. Otherwise, set rij = 1. The bit-map (rij ) is called the
rasterization of G in resolution w by h.
Suppose that i0 , j0 and j1 are integers such that 0 ≤ i0 < w and 0 ≤ j0 ≤
j1 < h. The triple (i0 , j0 , j1 ) is called a vertical segment in the rasterization of
G in resolution w by h if ri0 j = 1 for each integer j = j0 , . . . , j1 . The number
(j1 + 1 − j0 ) /h is considered to be the length of (i0 , j0 , j1 ).
18
WAYNE LEWIS AND PIOTR MINC
Observe that rasterizing a graphic G ⊂ I 2 in a high resolution w by h allows one to display a magnification of any part of G. Simply, take any i0 ∈
{0, . . . , w − wd − 1}, any j0 ∈ {0, . . . , h − hd − 1}, and set pij = ri0 +i j0 +j for
i = 0, . . . , wd − 1 and j = 0, . . . , hd − 1. This displays the rasterization
G ∩ {(x, y) | i0 /w ≤ x ≤ (i0 + wd ) /w and j0 /h ≤ y ≤ (j0 + hd ) /h}
on our screen.
Theorem 5.2. Suppose n, w and h are positive integers. Suppose also that J is
a subinterval of I and g : J → I is an n-crooked map. Suppose x0 , x1 ∈ J and µ
is an integer such that x0 < x1 and dwx1 e − bwx0 c ≤ 3µ . Then, the rasterization
of the graph of g in resolution w by h has a vertical segment of length at least
|g (x0 ) − g (x1 )| − (2µ + 2) /n.
Proof. Let m denote the minimum of g (x0 ) and g (x1 ), and let M denote the
maximum. Since the graph of g is nonempty, its rasterization in resolution w
by h has a vertical segment of length 1/h. So, we may assume that M − m =
|g (x0 ) − g (x1 )| > (2µ + 2) /n, because the theorem is true otherwise. Let k be the
least integer such that m ≤ k/n, and l be the greatest integer such that l/n ≤ M .
It follows that 0 ≤ k ≤ k + µ < l − µ ≤ n. By Proposition 3.22, there are numbers
a1 , b1 , a2 , b2 , . . . , a³3µ , b3µ such
´ that x0 ≤ a1 < b1 ≤ a2 < b2 ≤ · · · ≤ a3µ < b3µ ≤ x1
l−µ
and g (as , bs ) = k+µ
,
for s = 1, . . . , 3µ . Since dwx1 e − bwx0 c ≤ 3µ , the
n
n
set (x0 , x1 ) ∩ {1/w, 2/w, . . . , (w − 1) /w} has less than 3µ elements. Thus, there
is s0 ∈ {1, . . . , 3µ } such that (as0 , bs0 ) ∩ {1/w, 2/w, . . . , (w − 1) /w}
£ = ∅. ¤Let i0
be the greatest integer such that i0 /w ≤ as0 . Clearly, [as0 , bs0 ] ⊂ iw0 , i0w+1 . Let
j0 be the greatest integer such that j0 /h ≤ (k + µ) /n, and let j1 be the greatest
integer such that j1 /h ≤ (l − µ) /n. Observe that the triple (i0 , j0 , j1 ) is a vertical
segment in the rasterization of the graph of g in resolution w by h. The length of
(i0 , j0 , j1 ) is (j1 + 1 − j0 ) /h ≥ (l − µ) /n − (k + µ) /n = (l + 1) /n − (k − 1) /n −
(2µ + 2) /n. Since (l + 1) /n ≥ M and (k − 1) /n ≤ m, we have that (l + 1) /n −
(k − 1) /n ≥ M − m = |g (x0 ) − g (x1 )|. Consequently, the length of (i0 , j0 , j1 ) is at
least |g (x0 ) − g (x1 )| − (2µ + 2) /n.
¤
Remark 5.3. Observe that if m = k/n and M = l/n, then l − n − k/n =
|g (x0 ) − g (x1 )|. Consequently, in this case the vertical segment (i0 , j0 , j1 ) has
its length at least |g (x0 ) − g (x1 )| − 2µ/n.
Theorem 5.2 and Remark 5.3 imply the following corollary.
Corollary 5.4. Suppose n, w and h are positive integers, and g : I → I is an
n-crooked surjection. Suppose µ is an integer such that w ≤ 3µ . Then, the rasterization of the graph of g in resolution w by h has a vertical segment of length at
least 1 − 2µ/n.
¤
Theorem 5.2 implies that if we view the graph of an n-crooked function on
a raster device, then even for moderate choices of n, we must see long vertical
segments. For example, use Corollary 5.4 with n = 200 and µ = 50. Since 350 ≈
7.1 × 1023 , we cannot possibly take the horizontal resolution w greater than this
astronomical number. Therefore, the rasterized picture must contain straight black
vertical segments at least 1 − 2µ/n = 1/2 long where 1 is the height of the entire
graphic.
DRAWING THE PSEUDO-ARC
19
As we know from Section 4, for large n, the rasterized graph of σn consists of a
large black polygon with possibly some white pixels near the top and the bottom
(see Figures 5 and 6). Notice that Proposition 4.1 allows one to estimate the
distance of white pixels from the top and the bottom.
Rasterized graphs of arbitrary n-crooked surjections of I may look quite differently from the graph of σn . We construct two such examples below. However the
pictures must always contain long vertical arcs as promised by Theorem 5.2. We
construct n-crooked maps in Examples 5.6 and 5.7 by starting from the simple ncrooked map σn , and stretching some of the basic intervals then squeezing others.
We base these construction on the following simple observation.
Proposition 5.5. Suppose n is a positive integer and g is an n-crooked surjection
of I onto itself. Suppose also f is a continuous surjection of I onto itself. Then,
g ◦ f is an n-crooked surjection of I onto itself.
¤
Example 5.6. Let m and n be positive integers. Consider two increasing finite
sequences α = (a0 , a1 , a2 , . . . , am ) and β = (b0 , b1 , b2 , . . . , bm ) such that a0 = b0 = 0
and am = bm = 1. Let f : I → I be the piecewise linear map taking bi to ai for each
i = 0, . . . , m. Set g = σn ◦ f . By Proposition 5.5, g is n-crooked. Observe that by
taking consecutive terms ai growing very slowly from 0 and bi rapidly approaching
1, we can make, for several values of w, the rasterized graph of g to consist of one
horizontal row of pixels on the bottom and one vertical column on the right.
Example 5.7. For any positive integer k, let ∆k be the set of all sequences
Sl
δ1 δ2 . . . δk of 0’s and 1’s beginning with δ1 = 0. Let ∆l = k=1 ∆k .
Let n and m be positive integers such that 2m < n. Using the n-crookedness
of σn , one can find numbers x0 , y0 and z0 such that 0 < x0 < y0 < z0 < 1,
σn (y0 ) = 1/n and σn (x0 ) = σn (z0 ) = (n − 1) /n. Suppose that µ is an integer
such that 1 < µ ≤ m. Assume that, for each δ ∈ ∆µ−1 , xδ , yδ and zδ have been
constructed such that xδ < yδ < zδ , σn (yδ ) = (µ − 1) /n and σn (xδ ) = σn (zδ ) =
(n + 1 − µ) /n. Using the n-crookedness of σn , find xδ0 , yδ0 , zδ0 , xδ1 , yδ1 and zδ1
such that xδ < xδ0 < yδ0 < zδ0 < yδ < xδ1 < yδ1 < zδ1 < zδ , σn (yδ0 ) = σn (yδ1 ) =
µ/n and σn (xδ0 ) = σn (zδ0 ) = σn (xδ1 ) = σn (zδ1 ) = (n − µ) /n. Continue this
construction until xδ , yδ and zδ are constructed for all sequences δ ∈ ∆m .
Let C denote the standard ternary Cantor set. Let {Mδ | δ ∈ ∆k } be the collection, in the natural order, of theS
intervals removed in the k-th step of the construction C. Denote by M m the set δ∈∆m Mδ . Observe that the set C m = I \ M m is
the union of the intervals left after the m-th step of the construction of C.
Let f be the map of I onto itself such that f (0) = 0, f (1) = 1, f (Mδ ) =
{xδ } forSeach δ ∈ ∆m , and f is linear on each component of the complement of
{0, 1} ∪ δ∈∆m Mδ . Denote by g the composition σn ◦ f .
Finally let Inm = [m/n, (n − m) /n] and let G denote the graph of G. Observe
that (I × Inm ) ∩ G ⊂ C m × Inm . On the other hand, every point of C m × Inm is
within 3−m of some point of (I × Inm ) ∩ G. So, if we view (I × Inm ) ∩ G in several
horizontal resolutions, starting with small w and slowly increasing it to close to 3m ,
we get rasterized images similar to those of the Cantor set cross the interval Inm .
6. Rasterized images of hereditary indecomposable continua
In this section we observe that no black and white image can look hereditarily
indecomposable on a raster device. As in the previous section, we assume that
20
WAYNE LEWIS AND PIOTR MINC
our display device is tiled by a wd by hd grid of congruent square pixels Pij where
i = 0, . . . , wd − 1 and j = 0, . . . , hd − 1. Pij is the grid square in the xy-plane with
the left lower corner at (is, js) where s is the length of the sides. Let D denote the
collection of all pixels in the display device.
Definition 6.1. By the grid-distance between two pixels Pi0 j 0 and Pi00 j 00 we understand the maximum of |i0 − i00 | and |j 0 − j 00 |.
For any nonempty collection of pixels P ⊂ D we adopt the following definitions:
(1) Let d (P) denote the maximal grid-distance between two pixels in P. By
the grid-diameter of P we understand d (P) + 1.
(2) If k ≥ 0 is an integer, then (P, k) denotes the collection of those elements
of D whose grid-distance to some element of P is less than or equal to k.
(3) P is connected if its union is connected.
Proposition 6.2. Let P ⊂ D be a connected collection with grid-diameter equal n.
Let p denote the number of elements in P. Then n ≤ p ≤ n2 .
¤
Theorem 6.3. Let P ⊂ D be a connected collection with grid-diameter equal n.
Let p denote the number of elements in P. Suppose k and m are positive integers
such that (n − km) 2m > p. Then, there are two connected collections A0 ⊂ P and
A00 ⊂ P such that A0 ∩ A00 6= ∅, A0 \ (A00 , k − 1) 6= ∅ and A00 \ (A0 , k − 1) 6= ∅.
Proof. Observe that the condition (n − km) 2m > p implies that n − km > 0 and
consequently n − km − 1 ≥ 0.
Let R be any connected subcollection of P and let R0 and R00 be elements of
R with grid-distance equal to some integer ` ≥ k. Since R is connected there is
a finite sequence R0 , R1 , . . . , Rr of elements of R such that R0 = R0 , Rr = R00
and Ri−1 ∩ Ri 6= ∅ for each i = 1, . . . , r. Let q be the least integer such that the
grid-distance between Rq and R00 is k. Observe that the grid-distance between R0
and Rq is at least ` − k. Let hR, R0 , R00 i denote the collection {R0 , R1 , . . . , Rq }.
Clearly, hR, R0 , R00 i is connected, and R00 does not belong to (hR, R0 , R00 i , k − 1).
Notice that if hR, R0 , R00 i ∩ hR, R00 , R0 i 6= ∅ for some connected R ⊂ P, then the
theorem is true.
Now, suppose that the theorem is false and construct m+1 collections Q0 , Q1 , . . . ,
Qm such that, for each i = 0, . . . , m, Qi has the following properties:
• Qi is a collection of 2i mutually exclusive connected subcollections of P.
• The grid-diameter of each element of Qi is at least n − ik.
Set Q0 = {P}. Suppose that Qi−1 has been constructed for some i = 1, . . . , m.
Since n − km − 1 ≥ 0 and i ≤ m, n − (i − 1) k − 1 ≥ n − (m − 1) k − 1 ≥ k. For
each R ∈ Qi−1 choose R0 , R00 ∈ R with the grid-distance equal n − (i − 1) k − 1.
This choice of R0 and R00 together with some choice of R0 , R1 , . . . , Rr produces two
nonintersecting collections hR, R0 , R00 i and hR, R00 , R0 i. Let
Qi = {hR, R0 , R00 i , hR, R00 , R0 i | R ∈ Qi−1 } .
In other words, Qi is the collection that consists of 2i elements, grouped in 2i−1
pairs hR, R0 , R00 i and hR, R00 , R0 i corresponding to each R ∈ Qi−1 . Observe that
so constructed Qi has the properties promised above. Thus, the construction of
Q0 , Q1 , . . . , Qm is complete.
It follows that the union of Qm is a subcollection of P with at least (n − km) 2m
elements. But, (n − km) 2m is greater than the number of elements of P. This
contradiction completes the proof of the theorem.
¤
DRAWING THE PSEUDO-ARC
21
Definition 6.4. Suppose that P ⊂ D is a connected collection with grid-diameter
n. Let k be the greatest integer with the property that there are two connected
collections A0 ⊂ P and A00 ⊂ P such that A0 ∩ A00 6= ∅, A0 \ (A00 , k − 1) 6= ∅ and
A00 \ (A0 , k − 1) 6= ∅. By the relative decomposability number of P we understand
the ratio k/n. By the absolute decomposability number of P we understand the
product ks, where s is the length of the pixel sides.
Corollary 6.5. Each connected collection P ⊂ D with grid-diameter n has the
relative decomposability number greater than 1/ (1 + 2 log2 n) − 1/n.
Proof. Let p denote the number of elements in P. Clearly, p ≤ n2 . Also, if p = n2 ,
then the corollary is obvious. So, we may assume that p < n2 .
Set m = d2 log2 ne and k = dn/me − 1. Since k < n/m, (n − km) 2m ≥ 2m ≥
2
n > p. It follows from Theorem 6.3 that the relative decomposability number P is
at least k/n. But, k/n ≥ (n/m − 1) /n = 1/m − 1/n > 1/ (1 + 2 log2 n) − 1/n. ¤
Observation 6.6. A rasterized image of a continuum cannot look hereditarily indecomposable.
Explanation. Suppose P is the collection of black pixels in the rasterized image
of some continuum. Let n be the grid-diameter of P. For the sake of simplicity
assume that n is equal to the vertical resolution hd of the display device. If hd
is 103 (or less) then the relative decomposability number of P is at least 1/21 by
Corollary 6.5. This means that P contains two intersecting connected clusters of
pixels sticking out of each other by a well visible 1/21-st of the height of the display
device. Increasing hw = n to 106 does not do much. Corollary 6.5 guaranties that,
in that case, the absolute decomposability number of P must at least 1/41-st of
the height of the display device, still well visible. To gain some perspective on
these numbers suppose that the height of the display device is 10 inches. Then
hd = 103 makes the vertical resolution to be a moderate 100 dots per inch. In this
case the absolute decomposability number of P is estimated from below by about
0.5 in. Increasing hd to 106 decreases this last estimate only by a factor of 2, while
increasing the vertical resolution to 105 dpi, by far exceeding the resolution of the
human eye.
¤
Observation 6.6 can be extended to a magnified image of any part of a continuum.
Suppose that the intersection of a continuum C with a rectangle R is displayed on
a raster device. Then, by Observation 6.6, the rasterization of each component of
C ∩ R cannot look hereditarily decomposable.
7. Final comments
This paper addresses the possibilities of drawing the pseudo-arc or the graphs
of n-crooked maps, or of recognizing when a drawing represents the pseudo-arc or
a hereditarily indecomposable continuum. The results above show the limitations
on ability to draw the graph of an n−crooked map even for very modest sized n’s.
Thus one should expect that if the pseudo-arc or any nondegenerate hereditarily
indecomposable continuum appears naturally in a dynamical system, generating it
by plotting consecutive iterations of a map may give features suggestive of crookedness but will not fully show such. Also, the presence of sizable arcs in such a plot
should not necessarily be taken as evidence that the continuum is not hereditarily
indecomposable. The pseudo-arc results from an inverse limit, not from a single
22
WAYNE LEWIS AND PIOTR MINC
crooked piecewise linear map. As observed above, the size of maximal arcs can
decrease very slowly in such a construction as their number increases at a superastronomical rate.
The pseudo-arc is an inverse limit of arcs, but it can also be obtained as an
inverse limit of any nondegenerate connected polyhedron P. Though being arc-like,
a representation of it may look more like another polyhedron. In [25] the first author
used a construction of the pseudo-arc as an inverse limit of simple n-ods to show
that it admits period n homeomorphisms for each n. Thus a drawing which looks
simple n-od-like may still represent a pseudo-arc. Taking the n distinct images of
a subcontinuum not containing the fixed point of such a homeomorphism, one also
obtains n almost parallel copies of the continuum, which may also appear in any
representation of the pseudo-arc.
Much work on the pseudo-arc has made use of sequences of nested chains or
covers. Such have been useful not only in obtaining topological results but also
in dynamics [15], [16], [17], [32]. A construction using crooked chains or covers is
logically equivalent to a construction using inverse limits of arcs. However, the two
processes may well be perceived differently. Nested crooked chains readily suggest
an iterative process and do not suggest that one is after any stage illustrating the
actual continuum, but rather its method of construction. A subchain of a refining
chain passing without fold through two or three consecutive links of a containing
chain is unlikely to be interpreted as implying the existence of arcs. One can easily
concentrate on properties of subchains of refining chains without being concerned
with the superastronomical number of such subchains passing through a given part
of a containing chain. Once one realizes that links in a chain need not be connected,
in fact considering only the pseudo-arc and not the pseudo-arc in the plane links
cannot be connected, one can more readily visualize how every point of a pseudoarc is an endpoint. Amalgamating links and matching up chains and patterns gives
one a convenient way of constructing maps or homeomorphisms of pseudo-arcs,
building the pseudo-arcs and maps with given properties in the same process.
The pseudo-arc is an object with many interesting topological properties. Its
embeddings in the plane and what role it may play in dynamics are much less
well understood. It is convenient to understand what images of it generated by
various processes may look like and how to recognize such and the limitations of
such images.
References
1. R. D. Anderson and Gustave Choquet, A plane continuum no two of whose nondegenerate
subcontinua are homeomorphic: An application of inverse limits, Proc. Amer. Math. Soc. 10
(1959), pp. 347–353.
2. Marcy Barge and Joe Martin, The construction of global attractors, Proc. Amer. Math. Soc.
110 (1990), pp. 523–525.
3. R. H. Bing, A homogeneous indecomposable plane continuum, Duke Math. J. 15 (1948), pp.
729–742.
4. R. H. Bing, Concerning hereditarily indecomposable continua, Pacific J . Math. 1 (1951), pp.
43–51.
5. R. H. Bing, Higher-dimensional hereditarily indecomposable continua, Trans. Amer. Math.
Soc. 71 (1951), pp. 267-273.
6. R. H. Bing, Snake-like continua, Duke Math. J. 18 (1951), pp. 653–663.
7. Morton Brown, On the inverse limit of Euclidean n-spheres, Trans. Amer. Math. Soc. 96
(1960), pp. 129–134.
DRAWING THE PSEUDO-ARC
23
8. P. D. B. Collins, A. D. Martin, and E. J. Squires, Particle Physics and Cosmology, WileyInterscience, New York, 1989.
9. Eugene L. Fiume, The mathematical structure of raster graphics, Academic Press, Inc.,
Boston, MA, 1989.
10. Michael Handel, A pathological area preserving C ∞ diffeomorphism of the plane, Proc. Amer.
Math. Soc. 86 (1982), pp. 163–168.
11. George W. Henderson, The pseudo-arc as an inverse limit with one binding map, Duke Math.
J. 31 (1964), pp. 421–425.
12. M. Hénon, A two-dimensional mapping with a strange attractor, Comm. Math. Phys. 50
(1976), pp. 69–77.
13. Michael-R. Herman, Construction of some curious diffeomorphisms of the Riemann sphere,
J. London Math. Soc. 34 (1986), pp. 375–384.
14. J. R. Isbell, Embeddings of inverse limits, Ann. of Math. 70 (1959), pp. 73–84.
15. H. Kato, Continuum-wise expansive homeomorphisms, Can. J. Math. 45 (1993) pp. 576–598.
16. H. Kato, The nonexistence of expansive homeomorphisms of chainable continua, Fund. Math.
149 (1996), pp. 119–126.
17. Hisao Kato and Christopher Mouron, Hereditarily indecomposable compacta do not admit
expansive homeomorphisms, Proc. Amer. Math. Soc. 138 (2008), 3689–3696.
18. Judy A. Kennedy, A brief history of indecomposable continua, Continua (Cincinnati, OH,
1994), 103–126, Lecture Notes in Pure and Appl. Math., 170, Dekker, New York, 1995.
19. Judy A. Kennedy and James A. Yorke, Pseudocircles in dynamical systems, Trans. Amer.
Math. Soc. 343 (1994), pp. 349–366.
20. Judy A. Kennedy and James A. Yorke, Bizarre topology is natural in dynamical systems, Bull.
Amer. Math. Soc. (N.S.) 32 (1995), pp. 309–316.
21. Judy A. Kennedy and James A. Yorke, Pseudocircles, diffeomorphisms and perturbable dynamical systems, Ergodic Theory Dynam. Systems 16 (1996), pp. 1031–1057.
22. B. Knaster, Un continu dont tout sous-continu est indécomposable, Fund. Math. 3 (1922),
pp. 247–286.
23. J. Krasinkiewicz and P. Minc, Nonexistence of universal continua for certain classes of curves,
Bull. Acad. Polon. Sci. Sér. Sci. Math. Astronom. Phys. 24 (1976), pp. 733–741.
24. J. Krasinkiewicz and P. Minc, Mappings onto indecomposable continua, Bull. Acad. Polon.
Sci. Sér. Sci. Math. Astronom. Phys. 25 (1977), pp. 675–680.
25. Wayne Lewis, Periodic homeomorphisms of chainable continua, Fund. Math. 117 (1983), pp.
81–84.
26. Wayne Lewis, Observations on the Pseudo-Arc, Top. Proc. 9 (1984), pp. 329–337.
27. Wayne Lewis, The Pseudo-arc, Bol. Soc. Mat. Mexicana 5 (1999), pp. 25–77.
28. Brian F. Martensen, The local topological complexity of C r -diffeomorphisms with homoclinic
tangency, Nonlinearity 16 2003, pp. 161–186.
29. Piotr Minc and W. R. R. Transue, A transitive map on [0, 1] whose inverse limit is the
pseudoarc, Proc. Amer. Math. Soc. 111 (1991), pp. 1165–1170.
30. Edwin E. Moise, An indecomposable plane continuum which is homeomorphic to each of its
nondegenerate subcontinua, Trans. Amer. Math. Soc. 63 (1948), pp. 581–594.
31. Edwin E. Moise, A note on the pseudo-arc, Trans. Amer. Math. Soc. 67 (1949), pp. 57–58.
32. Christopher Mouron, Tree-like continua do not admit expansive homeomorphisms, Proc.
Amer. Math. Soc. 130 (2002), pp. 3409-3413.
33. S. B. Nadler, Jr., Continuum Theory: An Introduction, Marcel Dekker, Inc., New York, 1992.
34. Mark H. Turpin, The restricted rotation number of an example of Handel, Dynamical systems
and differential equations, Vol. II (Springfield, MO, 1996). Discrete Contin. Dynam. Systems
1998, Added Volume II, pp. 253–258.
35. Mark H. Turpin, An annulus diffeomorphism with non-Denjoy minimal sets, Proc. Amer.
Math. Soc. 126 1998, pp. 1851–1856.
24
WAYNE LEWIS AND PIOTR MINC
(Wayne Lewis) Department of Mathematics and Statistics, Texas Tech University,
Lubbock, Texas 79409
E-mail address: wayne.lewis@ttu.edu
(Piotr Minc) Department of Mathematics and Statistics, Auburn University, Auburn,
Alabama 36849
E-mail address: mincpio@auburn.edu