2.13 (a) If A and B are disjoint closed sets in some metric space X, prove that they are
separated.
Rudin’s Ex. 19
(b) Prove the same for disjoint open sets.
(c) Fix p ∈ X, δ > 0, define A to be the set of all q ∈ X for which d(p, q) < δ,
define B similarly, with > in place of <. Prove that A and B are separated.
(d) Prove that every connected metric space with at least two points is uncountable.
Hint: Use (c).
Proof (a) If A and B are closed, then A = A and B = B. By A ∩ B = ∅, we have
A ∩ B = A ∩ B = ∅ and A ∩ B = A ∩ B = ∅. Hence A and B are separated.
(b) Since A ∩ B = ∅, we know that B ⊂ Ac . The complement Ac is closed, since A
is open. Hence B ⊂ Ac = Ac . Thus A ∩ B ⊂ A ∩ Ac = ∅, which implies A ∩ B = ∅.
Similarly, we have A ∩ B = ∅. Therefore, A and B are separated.
(c) The sets A and B are
A = {x ∈ X : d(x, p) < δ},
B = {x ∈ X : d(x, p) > δ}.
The set A is open, by Theorem 2.19. To see that B is open, we let x ∈ B be any point.
By d(x, p) > δ, we take any number r such that 0 < r ≤ d(x, p) − δ. We claim that
Nr (x) ⊂ B. In fact, if z ∈ Nr (x), then d(z, p) ≥ d(x, p) − d(z, x) > d(x, p) − r ≥ δ,
which implies z ∈ B. Thus B is open.
It is clear that A and B are disjoint. Hence, by (b), A and B are separated.
(d) Suppose p and q are two distinct points of a connected metric space X. Denote
L = d(p, q) > 0. For each number r, with 0 < r < L, put
Ar = {x ∈ X : d(x, p) < r},
Br = {x ∈ X : d(x, p) > r}.
Since p ∈ Ar and q ∈ Br , we know that both sets are non-empty. By (c), we further
know that A and B are separated.
By the condition, X is connected, so X 6= Ar ∪ Br . Since Ar ∪ Br ⊂ X, there exists
a point zr such that zr ∈ X − (Ar ∪ Br ). By the definitions of Ar and Br , we know
that d(zr , p) = r.
If r 6= s, then d(zr , p) 6= d(zs , p), so zr 6= zs . Thus, r 7→ zr defines an injective map
from (0, L) to X. Since (0, L) is uncountable, so is X.
2.14 Let A and B be separated subsets of some Rk , suppose a ∈ A, b ∈ B, and define
p(t) = (1 − t)a + tb
for t ∈ R. Put A0 = p−1 (A), B0 = p−1 (B). [Thus t ∈ A0 if and only if p(t) ∈ A.]
(a) Prove that A0 and B0 are separated sets of R.
(b) Prove that there exists t0 ∈ (0, 1) such that p(t0 ) ∈
/ A ∪ B.
(c) Prove that every convex subset of Rk is connected.
1
Rudin’s Ex. 21
Proof (a) We shall prove A0 ∩ B0 = ∅.
In fact, if A0 ∩ B0 6= ∅, we let x ∈ A0 ∩ B0 . Then x ∈ A0 , and x ∈ B0 ∪ B00 . If
x ∈ B0 , then x ∈ A0 ∩ B0 , which implies p(x) ∈ A ∩ B ⊂ A ∩ B. This contradicts
to the assumption that A and B are separated. Thus, x is a limit point of B0 . We
shall prove that p(x) is a limit point of B.
For any neighborhood Nr (p(x)), we need to prove that Nr (p(x)) contains a point
p(y) in B, with p(y) 6= p(x). Indeed, notice that since A and B be separated,
hence, for a ∈ A and b ∈ B, we have a 6= b. Since x a limit point of B0 , in the
neighborhood of x:
(x − r/|b − a|, x + r/|b − a|) ,
there exists a point y 6= x such that y ∈ B0 , so p(y) ∈ B. From
|p(y) − p(x)| = |[(1 − y)a + yb] − [(1 − x)a + xb]| = |y − x| · |b − a|,
we know that p(y) 6= p(x), and p(y) ∈ Nr (p(x)). This concludes that p(x) is a limit
point of B, or p(x) ∈ B 0 . Thus, again we have p(x) ∈ A ∩ B, which contradicts to
the assumption that A and B are separated.
Therefore, A0 ∩ B0 = ∅. It is similar to prove that A0 ∩ B0 = ∅. Hence, A0 and B0
are separated.
(b) Suppose for every t ∈ (0, 1), we have p(t) ∈ A ∪ B. Since a = p(0) ∈ A and
b = p(1) ∈ B, we have p(t) ∈ A∪B for every t ∈ [0, 1]. This implies that t ∈ A0 ∪B0 .
Hence, [0, 1] ⊂ A0 ∪ B0 . It follows that
[0, 1] = {A0 ∩ [0, 1]} ∪ {B0 ∩ [0, 1]} .
It is clear that 0 ∈ A0 ∩ [0, 1], and 1 ∈ B0 ∩ [0, 1]. By (a), A0 and B0 are separated,
so we have
{A0 ∩ [0, 1]} ∩ {B0 ∩ [0, 1]} ⊂ A0 ∩ B0 = ∅,
and
[B0 ∩ (0, 1)] ∩ [A0 ∩ (0, 1)] ⊂ B0 ∩ A0 = ∅.
Hence, the segment [0, 1] is a union of two non-empty separated sets, but this contradicts to the fact that [0, 1] is connected, by Theorem 2.47.
Hence, there is a point t0 ∈ [0, 1] such that p(t0 ) ∈
/ A ∪ B. Apparently, t0 can be
neither 0 nor 1, since 0 ∈ A and 1 ∈ B. Therefore, there is a point t0 ∈ (0, 1) such
that p(t0 ) ∈
/ A ∪ B. This completes the proof of (b).
(c) Let S be a convex subset of Rk . If S is not connected, by the definition, there are
two non-empty separated sets A and B such that S = A ∪ B. By (b), there exists
t0 ∈ (0, 1) such that p(t0 ) ∈
/ A∪B. This contradicts to the fact that p(t) ∈ S = A∪B
for every t ∈ (0, 1) for the convex set S. Hence, S is connected.
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