Exercises for Section 1.2
1. Prove that the union of any (even infinite) number of open sets is open.
Prove that the intersection of two (and hence of finitely many) open
sets is open. Give a counterexample for the intersection of infinitely
many open sets.
U
U
S
Proof. Let
be any family of open sets. Let U =
be the union
of all the sets in . We need to show that for each point x ∈ U there
is an open rectangle, B x with x ∈ B x and B x ⊂ U.
U
U
Choose x ∈ U. Then there is some open set V ∈
with x ∈ V . Since
x
V is open, there is an open rectangle B with x ∈ B x and B x ⊂ V .
But V ⊂ U, therefore B x ⊂ U.
Now we wish to show that given two open sets, A and B, their intersection A ∩ B is also open. To do this, we must first show that the
intersection of two open rectangles is open. Let C and D be open rectangles in Rn . If C ∩ D = ? then it is open and we are done. If they
do intersect, then the corresponding intervals forming each of their n
sides must intersect in Rn . But the non-empty intersection of two open
intervals is an open interval. Therefore, the resulting object is an open
rectangle.
Now let x ∈ A ∩ B. Then there are open rectangles C x and D x around
x with C x ⊂ A and D x ⊂ B. Their intersection is not empty (both
contain x), so C x ∩ D x is an open rectangle. Furthermore, it is clear
that C x ∩ D x ⊂ A ∩ B, so it proves that A ∩ B is open.
To see that infinitely many open sets do not, in general, intersect in an
open set, consider En = (−1/n, 1/n) for n ∈ N. Each of these sets is
open, but their intersection is just {0}, a closed set. Other examples
abound.
2. If A ⊂ B ⊂ Rn , prove that
clA ⊂ clB,
and
intA ⊂ intB.
Proof. First to show that clA ⊂ clB. Since A ⊂ B, we have for free
that A ⊂ clB = B ∪ ∂B. Now it suffices to show that ∂A ⊂ clB. Let
x ∈ ∂A. Let U x be any open set containing x. Then U x ∩ A 6= ?. But
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since A ⊂ B, we may conclude that U x ∩ B 6= ? as well. Therefore, x
is not in the exterior of B, so it must be in clB.
Now to show that intA ⊂ intB. This works on the same principle.
Let x ∈ intA. Then there is some open rectangl;e U x containing x
with U x ⊂ A. But A ⊂ B implies that U x ⊂ B also. Therefore,
x ∈ intB.
3. Prove that if B is an open subset of A, then B ⊂ int(A). Note that
this says that int(A) is the largest open subset of A.
Proof. Let x ∈ B (and therefore, x ∈ A also). Then since B is open,
there is an open rectangle U x with x ∈ U x ⊂ B. Since B ⊂ A, U x ⊂ A.
Therefore, x is an interior point, and B ⊂ intA.
4. Prove that the n-dimensional ball centered at a of radius r, B n (a; r) =
{x ∈ Rn : kx − ak < r} is open.
Proof. The entire idea behind the proof in general is captured by showing that B n (0; r) (the open ball of radius r centered at the origin) is
open. Let x ∈ B n (0; r). Then x = (x1 , . . . , xn ) and kxk < r. Let d
denote the distance r − kxk to the boundary of the ball. Then define
the open rectangle:
d
d
d
d
n
1
n
1
×···× x − √ ,x + √
.
A = x − √ ,x + √
2 n
2 n
2 n
2 n
√
Then A is an n-dimensional open rectangle of side length d/ n. And
every point in A is within a distance of d/2 from x. Therefore, all of A
is contained within B n (0; r), and the ball is open.
5. Find the interior, exterior, and boundary of the sets:
B n = {x ∈ Rn : kxk ≤ 1} ,
S n−1 = {x ∈ Rn : kxk = 1} ,
Qn = x ∈ Rn : each xi is rational .
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Solution.
intB n = {x ∈ Rn : kxk < 1}
(the open ball),
extB n = {x ∈ Rn : kxk > 1} ,
and
∂B n = S n−1 = {x ∈ Rn : kxk = 1}
(the unit sphere in Rn ).
intS n−1 = ?,
extS n−1 = {x ∈ Rn : kxk < 1} ∪ {x ∈ Rn : kxk > 1} ,
and
∂S n−1 = S n−1.
intQn = ?, extQn = ?, and ∂Qn = Rn (because every open rectangle
has points with rational coordinates).
6. If A ⊂ [0, 1] is the union of open intervals (ai , bi ) such that each rational
number in (0, 1) is contained in some (ai , bi ), show that ∂A = [0, 1]−A.
Proof. What we are really trying to prove here is that A and ∂A are
disjoint, and their union is [0, 1]. In other words, if x ∈ [0, 1] then x
belongs to exactly one of the two sets, A or ∂A. First, a couple of
remarks about the set A: (1) A is an open set because it is the union
of open intervals; and (2) every rational number in (0, 1) belongs to A.
By the first remark, we see that none of the points in A are in ∂A,
because A is open, so A = intA. All that remains is to show that
all other points in [0, 1] are in ∂A. Let x ∈ [0, 1] − A. Let U x be any
interval around x. Then U x must have rational numbers in it (rationals
are dense in R). Therefore, U x ∩ A 6= ?. On the other hand, x ∈ U x ,
so U x ∩ (R − A) 6= ?. Thus we have x ∈ ∂A.
7. If A is a closed set that contains every rational number r ∈ [0, 1], show
that [0, 1] ⊂ A.
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Proof. Let x ∈ [0, 1], and let U x be any open interval containing x.
Then U x must have rational numbers that lie within [0, 1]. Therefore,
U x intersects A. This means that x is not an exterior point of A. The
only possibility, then, is that either x ∈ ∂A or x ∈ intA. But since A
is closed, this implies that x ∈ A.
8. Graph generic open balls in R2 with respect to each of the “nonEuclidean” norms, k · k1 and k · k∞ . What shapes are they?
Solution. They are diamonds and squares, respectively.
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