CHAPTER 1
Continuity and convergence in Euclidean spaces
This section serves as motivation for the definitions of a topological space X and a
continuous function f : X → Y between two topological spaces X and Y , both given
in section ??. These definitions are straightforward but upon first encounter seem ad
hoc and arbitrary. To conway a sense of where they come from, we first reexamine
the familiar setting of continuous functions f : Rn → Rm in terms of the usual δ, ε
language from calculus or analysis. We then introduce the notion of open and closed
subsets of Euclidean spaces and use these to reinterpret continuity of functions on
Euclidean spaces. The main results of this section are stated in theorems 3.1 and 3.2.
1. Euclidean space
The n-dimensional Euclidean space Rn is defined as the set of ordered n-tuples of
real numbers:
Rn = {(x1 , ..., xn ) | xi ∈ R }
In the above, and throughout the text, the symbol R denotes the set of real numbers.
We will adopt the convention of denoting the n-tuple (x1 , ..., xn ) simply by x, the ntuple (y1 , ..., yn ) by y, etc. Two elements x = (x1 , ..., xn ) and y = (y1 , ..., yn ) from Rn
can be added and multiplied by a real number λ ∈ R in the familiar ways:
x + y = (x1 + y1 , ..., xn + yn )
and
λ · x = (λ · x1 , ..., λ · xn )
n
These operations endow R with the structure of a real, n-dimensional vector space.
Definition 1.1. For a real number p ≥ 1 and for p = ∞, we define the functions
dp : Rn × Rn → [0, ∞i as
 p
 p |x1 − y1 |p + ... + |xn − yn |p
;
if p =
6 ∞
(1)
dp (x, y) =

max {|x1 − y1 |, ..., |xn − yn |}
;
if p = ∞.
When p = 2 we call the resulting function d2 the Euclidean distance function and denote
it simply by d.
For example
d1 (x, y) = |x1 −y1 |+· · ·+|xn −yn |
and
3
4
d 3 (x, y) = |x1 − y1 | + · · · + |xn − yn |
4
while d = d2 takes on the familiar form
p
d(x, y) = d2 (x, y) = (x1 − y1 )2 + · · · + (xn − yn )2
1
3
4
43
2
1. CONTINUITY AND CONVERGENCE IN EUCLIDEAN SPACES
Definition 1.2. Let r > 0 be a real number and let x ∈ Rn be a point in Euclidean
space. We define the open Euclidean ball Bx (r) of radius r and center x as the subset
of Rn given by
Bx (r) = {y ∈ Rn | d(x, y) < r}
Note that in our definition of Bx (r) used the Euclidean distance function d. We
could equally well have used any of the distance functions dp (including the case of
p = ∞) in this definition. Thus, we’ll let Bxp (r) denote the open ball with center x and
radius r but taken with respect to the metric dp :
Bxp (r) = {y ∈ Rn | dp (x, y) < r}
p
Figure 1 illustrates the shapes of B(0,0)
(1) ⊂ R2 for the choices of p = 1, 2, ∞.
1
B(0,0)
(1)
2
B(0,0)
(1) = B(0,0) (1)
∞
B(0,0)
(1)
p
Figure 1. Examples of the open balls B(0,0)
(1) ⊂ R2 for p = 1, 2, ∞.
The dotted lines indicate that the boundaries of these shapes are not
part of the balls.
Definition 1.3. A function f : Rn → Rm is continuous at the point x ∈ Rn if for
every ε > 0 one can find a δ > 0 such that if d(x, y) < δ then d(f (x), f (y)) < ε. Said
differently, f is continuous at x ∈ Rn if for every ε > 0 there exists a δ > 0 such that
f (Bx (δ)) ⊆ Bf (x) (ε)
The function f is said to be continuous if it is continuous at every point x ∈ Rn .
Figure 2 illustrates this definition. Continuity of functions is closely tied to convergence of sequences. We briefly review this connection.
Definition 1.4. A sequence xk ∈ Rn is said to converge to x ∈ Rn if for every
ε > 0 there exists a positive integer k0 such that for all k ≥ k0 one obtains d(xk , x) < ε.
In this case we write limk→∞ xk = x or simply lim xk = x.
The following theorem is a standard result discussed in calculus and analysis courses,
its proof is omitted.
2. OPEN AND CLOSED SUBSETS OF EUCLIDEAN SPACE
f
3
Bf (x) (ε)
f (x)
x
Bx (δ)
Rn
Rm
Figure 2. The function f is continuous at x ∈ Rn if for every ε > 0
there is a δ > 0 such that the ball of radius δ centered at x (left shaded
disk) maps to the ball of radius ε > 0 with center f (x) (shaded disk on
the right). The amoeba-like shape inside of Bf (x) (ε) indicates the image
of Bx (δ) under f .
Theorem 1.5. Let f : Rn → Rm be a function and x ∈ Rn be any point. Then f is
continuous at x if and only if for every seqnce xk ∈ Rn that converges to x in Rn , the
sequence yk = f (xk ) converges to y = f (x) in Rm . Said differently, f is continuous at
x if and only if it commutes with the limit symbol lim when applied to any sequence xk
with limit x:
f lim xk = lim f (xk )
k→∞
k→∞
2. Open and closed subsets of Euclidean space
Given a set X and a subset A ⊂ X, we define the complement of A in X as the set
X − A = {x ∈ X | x ∈
/ A}
Definition 2.1. A subset A ⊆ Rn is called closed if for every convergent sequence
ai ∈ A the limit limi→∞ ai also lies in A. A subset B ⊆ Rn is called open if Rn − B is
closed.
Example 2.2. When n = 1, all closed intervals [a, b] are closed sets while open
intervals ha, bi are open sets in R. The sets R and ∅ are both closed and open while the
intervals [a, bi are neither closed nor open. Any finite subset of R is closed.
Example 2.3. When n = 2 examples of closed sets are the closed rectangle [a, b] ×
[c, d], the closed circles {(x, y) ∈ R2 | x2 + y 2 ≤ r}, the upper half-plane {(x, y) ∈
R2 | y ≥ 0}, the graph Γf of a continous function f : R → R: Γf = {(x, f (x)) | x ∈ R}.
The main examples of open sets are the open balls Bx (r).
4
1. CONTINUITY AND CONVERGENCE IN EUCLIDEAN SPACES
Proposition 2.4. A subset U ⊆ Rn is open if and only if for every x ∈ U there
exists a ε > 0 such that Bx (ε) ⊆ U .
Proof. =⇒ Let U be an open subset of Rn and let x ∈ U be an arbitrary
element. Suppose there were no ε > 0 for which Bx (ε) ⊆ U . In particular, for any
positive integer i and setting ε = 1/i, the ball Bx (1/i) would have to intersect the
complement V = Rn − U of U . Pick an arbitrary element bi ∈ V ∩ Bx (1/i), one for
each i = 1, 2, 3, .... thus creating a sequence bi ∈ V . The property d(x, bi ) < 1/i implies
that bi converges to x. However, since U is open, V must be closed and so x = lim bi
must lie in V . But clearly x ∈
/ V , creating a contradiction. Therefore, an ε with the
property Bx (ε) ⊆ U must exist.
⇐= To prove that U is open we need to prove that V = Rn − U is closed.
Supposed that this fails. Then there must exists a convergent sequence bi ∈ V whose
limit x = lim bi lies in U . Pick an ε > 0 so that Bx (ε) ⊆ U . Since none of the bi lie
in U , we see that d(bi , b) ≥ ε. This however is impossible for a convergent sequence,
creating a contradiction. We are then forced to conclude that V is a closed subset of
Rn and hence that U is an open set.
Example 2.5. For any value of n, any choice of x ∈ Rn and any choice of r > 0, the
set Bx (r) is open. This follows easily from the preceding proposition. For if y ∈ Bx (r),
let ρ = r−d(x, y), note that ρ is positive. An easy check then shows that By (ρ) ⊂ Bx (r).
3. Continuity and convergence in terms of open and closed sets
The purpose of this section is to prove that the continuity of a function f : Rn → Rm
either at a point or globally, can be expressed entirely in terms of open or closed sets.
Recall that, given a function f : X → Y between two sets X and Y , the preimage
−1
f (V ) of a subset V ⊂ Y is defined as
f −1 (V ) = {x ∈ X | f (x) ∈ V }
In particular, f −1 (V ) is a subset of X.
Theorem 3.1. Let f : Rn → Rm be a function.
(a) The function f is continous at x ∈ Rn if and only if for every open set V ⊆ Rm
containing f (x), there exists an open set U ⊆ Rn containing x with the property
that f (U ) ⊂ V .
(b) The function f is continuous if and only if for every open set V ⊆ Rm the
preimage U = f −1 (V ) of V under f is open in Rn .
Proof. (a) =⇒ Assume that f is continuous at x ∈ Rn and let V ⊂ Rm be any
open set containing f (x). Since V is open, by proposition 2.4, there exists an ε > 0
such that Bf (x) (ε) is contained in V . But continuity of f at x then implies the existence
of a δ > 0 such that f (Bx (δ)) ⊂ Bf (x) (ε). Since Bx (δ) is an open set (see example 2.5)
and since Bf (x) (ε) is contained in V , by taking U = Bx (δ), we see that f (U ) ⊂ V , as
claimed, see figure 3.
3. CONTINUITY AND CONVERGENCE IN TERMS OF OPEN AND CLOSED SETS
f
5
V
Bf (x) (ε)
Bx (δ)
x
Rn
Rm
Figure 3. This picture illustrates the portion (a) =⇒ of the proof of
theorem 3.1. V is an arbitrary open set containing f (x) and ε > 0 is
chosen so that Bf (x) (ε) (shaded disk on the right) is contained in V . By
continuity of f at x, there is a δ > 0 so that Bx (δ) (shaded disks on
the left) maps into B(f (x) (ε) under f . The image of Bx (δ) under f is
indicated by the smaller amoeba-like shape inside of Bf (x) (ε).
⇐= Assume that f has the property that for every open subset V ⊂ Rm that
contains f (x) there is an open subset U ⊂ Rn containing x such that f (U ) ⊂ V . We’d
like to show that then f is continuous at x. Thus pick an arbitrary ε > 0 and let
V = Bf (x) (ε). Note that V is an open set containing f (x) and so we are guaranteed
the existence of an open set U containing x such that f (U ) ⊂ Bf (x) (ε). But since U
is open, there must exist a δ > 0 such that Bx (δ) ⊂ U (by proposition 2.4). Thus we
found a δ such that f (Bx (δ)) ⊂ Bf (x) (ε) and so f is continuous at x, see figure 4.
(b) =⇒ Suppose f is continuous and let V ⊂ Rm be any open set. We like to show
that U = f −1 (V ) is also an open set. Let x ∈ U be any point. Since f is continuous
at x, part (a) of the theorem shows that there exists an open set Ux with f (Ux ) ⊂ V .
In particular, Ux ⊂ U . But since Ux is open and x ∈ Ux , there must exist a δ > 0 so
that Bx (δ) ⊂ Ux and so Bx (δ) ⊂ U . According to proposition 2.4, this shows that U is
open.
⇐= Suppose that f has the property that f −1 (V ) is an open set whenever V is an
open set. We’d then like to show that f must be continuous. By definition, this means
that we must show that f is continuous at each point x ∈ Rn . Using part (a) of the
theorem, demonstrating the continuity of f at x is equivalent to showing that for each
open V ⊂ Rm containing f (x), there exists an open set U ⊂ Rn containing x and such
that f (U ) ⊂ V . But our working assumption allows us to simply take U = f −1 (V ),
thus completing the proof of the theorem.
6
1. CONTINUITY AND CONVERGENCE IN EUCLIDEAN SPACES
f
V = Bf (x) (ε)
U
Bx (δ)
x
Rn
Rm
Figure 4. This picture illustrates the portion (a) ⇐= of the proof of
theorem 3.1. Given any ε > 0 we set V = Bf (x) (ε) rendering it an open
set containing f (x). By assumption, there is an open set U (amoeba-like
set on the left) containing x and such that f (U ) ⊂ V . But since U is
open, there is a δ > 0 so that Bx (δ) (shaded disk on the left) is contained
in U . The images of U and Bx (δ) are indicated inside of V .
The importance of the preceding theorem is that it provides a definition of continuity that only relies on the notion of open sets. This observation will serve as the basis
for the definition of a topological space (section ?? below) and the generalization of
continuity to such settings. The next theorem testifies that one can in fact also define
continuity in terms of closed sets only.
Theorem 3.2. A function f : Rn → Rm is continuous if and only if f −1 (B) is a
closed subset of Rn whenever B is a closed subset of Rm .
Proof. =⇒ Suppose that f is continuous and let V ⊂ Rm be a closed set. We’d
like to show that A = f −1 (B) is a closed subset of Rn . To see this, note that
f −1 (Rm − B) = Rn − f −1 (B) = Rn − A
Since Rm − B is open and since according to theorem 3.1 f −1 (Rm − B) then also must
be open, the above equality of sets shows that Rn − A is open. By definition, this
means that A is closed.
⇐= Suppose that f −1 (B) is closed whenever B is closed. Let V ⊂ Rm be any
open set, let U = f −1 (V ) and set B = Rm − V (note that B is closed). Then, on one
hand, A = f −1 (B) is closed while on the other,
f −1 (B) = f −1 (Rm − V ) = Rn − f −1 (V ) = Rn − U
Thus U must be open and so, according to theorem 3.1, f must be continuous.
4. SOME PROPERTIES OF OPEN AND CLOSED SETS IN EUCLIDEAN SPACE
7
We next turn to sequences where, as with continuity, convergence can be expressed
solely in terms of open sets.
Theorem 3.3. A sequence xk ∈ Rn converges to x ∈ Rn if and only if for every
open set U ⊂ Rn containing x, there exists a natural number k0 such that for all k ≥ k0
we obtain xk ∈ U .
Proof. =⇒ Suppose that lim xk = x and that U is an open set containing x.
Then there exists an ε > 0 such that Bx (ε) ⊂ U . Find a natural number k0 such that
for all k ≥ k0 the points xk lie in Bx (ε). Clearly then xk ∈ U also for all k ≥ k0 .
⇐= Suppose now that xk is a sequence in Rn with the property that for every
open set U containing x, there is a natural number k0 such that k ≥ k0 implies that
xk ∈ U . Given any ε > 0, we need to demonstrate that there is a k0 so that k ≥ k0
implies that xk ∈ Bx (ε). Of course, choosing U = Bx (ε) finishes the proof (see example
2.5).
4. Some properties of open and closed sets in Euclidean space
In the proof of theorem 4.1 below, we shall use the following formulas knows as
DeMorgan’s laws. Let X be any set and let Ui ⊆ Rn be a family of subsets of X with
i running through some indexing set I. Then the following equalities of sets hold:
X − (∩i∈I Ui ) = ∪i∈I (X − Ui )
X − (∪i∈I Ui ) = ∩i∈I (X − Ui )
Said differently, these relations imply that
The complement of the intersection is the union of the complements.
The complement of the union is the intersection of the complements.
With these preliminaries in place, we are ready to turn to the main result of this
section.
Theorem 4.1. The following are properties of open and closed subsets of Euclidean
space:
(1) The sets Rn and ∅ are both open and closed sets.
(2) The union of any number of open sets is again an open set. The intersection
of any finite number of open sets is an open set.
(3) The intersection of any number of closed sets is again a closed set. The union
of any finite number of closed sets is a closed set.
Proof. Point (1) above follows directly from the definitions of open and closed
subsets of Euclidean space. To verify point (2), let Ui ⊆ Rn , i ∈ I be a family of open
sets and set U = ∪i Ui . We need to show that U is again an open set. Let x ∈ U .
Then x ∈ Ui for some i ∈ I. By proposition 2.4, there must exist a ε > 0 such that
Bx (ε) ⊆ Ui . But then Bx (ε) ⊂ U since Ui ⊆ U . This shows that for every x ∈ U there
is a ε > 0 such that Bx (ε) ⊆ U . According to proposition 2.4 this means that U is an
open set.
8
1. CONTINUITY AND CONVERGENCE IN EUCLIDEAN SPACES
On the other hand, let U1 , ..., Um be a finite family of open sets and let V = ∩m
j=1 Uj .
To see that V is open, pick an arbitrary x ∈ V . Then x ∈ Ui for every j ∈ {1, 2, ..., m}
and so there exist numbers εj > 0 with the property that Bx (εj ) ⊆ Uj . Let ε =
min{ε1 , ..., εm }. Clearly then Bx (ε) ⊆ Uj for every j ∈ {1, ..., m} so that Bx (ε) ⊆ V .
This shows that V is an open set.
For point (3) follows immediately from (2) by sing DeMorgan’s laws. Let Vi , i ∈ I
be a family of closed sets and let Ui = Rn − Vi . Then all the sets Ui are open and so
U = ∪i Ui is also open (by part (2) of the theorem). By DeMorgan’s second law we
know that
Rn − U = Rn − (∪i Ui ) = ∩i (Rn − Ui ) = ∩i Vi
Since U is open the set Rn − U is closed and so the above says that ∩i Vi is closed.
The part of (3) about finite unions of closed sets follows in a similar manner (using
DeMorgan’s first law) and is left as an easy exercise.
There are many examples of infinite families of open sets whose intersection is not
an open set as well as infinite families of closed sets whose union is not a closed set.
Here are two instances:
Example 4.2. For each i = 1, 2, 3, ... let
1 1
Ui = − ,
i i
Then each set Ui is open set but the intersection ∩∞
i=1 Ui = {0} is closed (but not open).
Example 4.3. For each i = 1, 2, 3, ... let
1
1
Vi = , 3 −
i
i
Note that each Vi is closed but their union ∪∞
i=1 Vi = h0, 3i is open (but not closed).