Math 5070 Fall 2012 Homework 4 solutions
1. 38.5 Let X be subseet of a metric space M . We say that a point x in
M is an accumulation point of X there exists a sequence {xn } in X such that
limx→∞ xn = x and xn 6= x for all n. Denote by X a the set of all accumulation
points of X.
(a) Prove that X is closed if and only if X a ⊂ X.
(b) Prove that if X is a bounded infinite subset of R, then X a 6= ∅.
(c) Prove that if X is an uncountable subset of R, then X a 6= ∅.
Solution.
(a) We have defined that X is closed if limn→∞ xn ∈ X for every sequence
{xn } in X which converges in M .
=⇒: Let x be an accumulation point of X. Then there exists a sequence
{xn } in X such that limx→∞ xn = x in M , and because X is closed, x ∈ X.
⇐=: Suppose that X a ⊂ X, {xn } is a sequence in X, and limn→∞ xn = x.
We need to show that x ∈ X. If xn = x for some n, then x ∈ X because xn ∈ X.
If xn 6= x for all n, then x ∈ X a ⊂ X.
(b) Because X is infinite, we can choose a sequence {xn } in X with all
xn different. Because X is bounded, the sequence {xn } is bounded, and by
the Weierstrass theorem, there is a convergent subsequence xnk → x. We will
show that x ∈ X a . If xnk 6= x for all k, we are done. Otherwise, since all
xnk are different, xnk = x for only one k, and by removing this term from the
subsequence we get a subsequence which of numbers from X all different from
x, which also converges to x.
Another solution. Because X is bounded, X ⊂ (a0 , b0 ) for some real a0 , b0 .
Then X ∩ (a0 , b0 ) is infinite. If we already have ak , bk such that X ∩ (ak , bk ) is
inifinite, at least one of the sets
ak + bk
ak + bk
, X∩
, bk
X ∩ ak ,
2
2
k
k
or ak +b
is infinite. Choose (ak+1 , bk+1 ) to be one of the intervals ak , ak +b
, bk
2
2
such that X ∩ (ak , bk ) is infinite. By induction, we get intervals
(a0 , b0 ) ⊃ (a1 , b1 ) ⊃ · · · ⊃ (ak , bk ) ⊃ · · ·
of lengths
b0 − a0
→ 0, k → ∞.
2k
Since ak is increasing and bounded (by b0 ), it has a limit L. Because limk→∞ bk −
ak = 0, bk has the same limit L. Since the set X ∩ (ak , bk ) has at least two
elements (in fact, it is infinite), we can choose xk ∈ X ∩ (ak , bk ), xk 6= L, and
by the squeeze theorem, limk→∞ xk = L.
(c) We first show that if x ∈ X but x ∈
/ X a then
bk − ak =
∃εx > 0 : (x − εx , x + εx ) ∩ X = {x} .
1
(1)
Otherwise
∀ε > 0 : (x − ε, x + ε) ∩ X 6= {x} ,
Because x ∈ (x − ε, x + ε) ∩ X always and choosing ε = 1/n in turn we get
∀n ∈ N∃xn ∈ (x − 1/n, x + 1/n) ∩ X, xn 6= x.
Since xn → x, n → ∞, and xn =
6 x for all n, it follows that x ∈ X a .
Taking the εx from (1), we have
[
X⊂
(x − εx /2, x + εx /2)
x∈X
and the sets (x − εx /2, x + εx /2) are disjoint; if
(x − εx /2, x + εx /2) ∩ (y − εy /2, y + εy /2) 6= ∅
for some x, y ∈ X, x 6= y, then |x − y| < εx /2+εy /2 ≤ max {εx , εy } and so either
y ∈ (x − εx , x + εx )∩X or x ∈ (y − εy , y + εy )∩X, which is a contradiction. By
the density of rationals, there exists a rational number rx ∈ (x − εx /2, x + εx /2)
and since the intervals are disjoint, rx is different for different x, and x 7→ rx is
a one to one mapping between X and a subset of rationals, which is countable.
Alternative solution. Define Xn = X ∩ [n, n + 1].If all Xn were finite, then
∞
[
X =
Xn is countable as the countable union of finite sets. Thus there
n=−∞
exists Xn which is infinite. Because Xn is bounded, ∅ 6= Xna and because
Xna ⊂ X a , it holds that X a 6= ∅.
2. 38.7 Let δ (k) be the sequences
δ (1) = (1, 0, 0, . . .)
δ (2) = (0, 1, 0, . . .)
..
.
n
o∞
(k)
(k)
(k)
that is, for each k, δ (k) is the sequence δn
defined by δk = 1, δn = 0
n=1
if n 6= k. Prove that the set X = δ (k) |k ∈ N is closed subset of `1 , `2 , c0 , and
`∞ .
Solution. First, X is a subset of all those spaces:
∞ X
(k) δn = 1 =⇒ δ (k) ∈ `1
n=1
∞ X
(k) 2
δn = 1 =⇒ δ (k) ∈ `2
h
n=1 i
∀n ∈ N : δn(k) ≤ 1 =⇒ δ (k) ∈ `∞
lim δn(k) = 0 =⇒ δ (k) ∈ c0
n→∞
2
(k)
where the last statement follows from the fact that the sequence δn = 0 for all
n > k.
Now, if i 6= j, then
d`1 δ (i) , δ (j) = 1 + 1 = 2
√
√
d`2 δ (i) , δ (j) = 1 + 1 = 2
d`∞ δ (i) , δ (j) = 1
Since always d δ (i) , δ (j) ≥ 1 for i 6= j, the only sequences from X that are
convergent are those that are eventually constant, and so X has no accumulation
points. The result follows from problem 1(a).
3. 38.8 Let X and Y be closed subsets of R. Prove that X × Y is a closed
subset of R2 . State and prove a generalization to Rn .
Solution. Consider X1 , . . . , Xn ⊂ R closed. The metric in Rn is given by
v
u n
uX
2
d ((x1 , . . . , xn ) , (y1 , . . . , yn )) = t
|xi − yi | .
i=1
Suppose x(k) is a sequence in X1 × · · · × Xn such that limk→∞ x(k) = x in
Rn , that is
lim d x(k) , x = 0.
k→∞
Then for all i = 1, . . . , n,
(k)
xi − xi ≤ d x(k) , x → 0 as k → ∞,
(k)
thus limk→∞ xi = xi in R. Since Xi are closed, xi ∈ Xi for all i = 1, . . . , n,
and, consequently,
x = (x1 , . . . , xn ) ∈ X1 × · · · × Xn .
4. 38.13 Let M be a metric space. Prove the following:
(a) X = X for all X ⊂ M
(b) X is closed for all X ⊂ M
(c) For all X, Y ⊂ M , if X ⊂ Y ⊂ M , then X ⊂ Y
(d) X ∪ Y = X ∪ Y for all X, Y ⊂ M
(e) If Y is a closed subset\
of M such that X ⊂ Y , then X ⊂ Y.
{Y |X ⊂ Y ⊂ M, Y is closed}
(f) If X ⊂ M , then X =
Solution.
(a) Since any set is a subset of its closure, we have X ⊂ X.We need to prove
X ⊂ X. Let x ∈ X. Then there exists a sequence {xn } ⊂ X, xn → x as
n → ∞, . Let m ∈ N. Since xn → x as n → ∞, there exists some n such that
d (xn , x) < 1/2m. Since xn ∈ X, there exists a sequence {yk } ⊂ X, yk → xn as
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k → ∞, and so there exists k such that d (yk , xn ) < 1/2m.Put zm = yk . Then
zm ∈ X and
d (zm , x) ≤ d (zm , xn ) + d (xn , x) <
1
1
1
+
= .
2m 2m
m
We have constructed a sequence {zm } ⊂ X, zm → x as m → ∞, so x ∈ X.
(b) Since X = X by (a), X is closed.
(c) If x ∈ X, then x = limn→∞ xn for some {xn } ⊂ X. Since X ⊂ Y, also
{xn } ⊂ Y , so x ∈ Y .
(d) Since X ⊂ X ∪ Y and Y ⊂ X ∪ Y , we have by (c) X ⊂ X ∪ Y and
Y ⊂ X ∪ Y , so X ∪ Y ⊂ X ∪ Y . For the opposite inclusion, let x ∈ X ∪ Y .
Then x = limn→∞ xn , where all xn ∈ X ∪ Y . Either infinitely many xn ∈ X,
in which case x ∈ X because the subsequence of xn ∈ X convergese to x, or
infinitely many xn ∈ Y and then x ∈ Y (or both).
(e) Since Y is closed, Y = Y . By (c), X ⊂ Y gives X ⊂ Y = Y .
(f) Denote F = {Y |X ⊂ Y ⊂ M, Y is closed}, the family of all closed sets
in M which contain X. Since all
\ sets in F are closed, and the intersection of a
family of closed sets is closed, F is closed. Since X is subset of any set Y in
\
\
F, it is contained in their intersection, so X ⊂ F. By (e), X ⊂ F. But
\
X ⊂ X and X is closed by (b), so X ∈ F, hence F ⊂ X.
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