Suggested solutions for homework I
1. Prove that every compact subset of a metric space is closed and bounded. Prove that a
closed subset of a compact space is compact.
Proof. (i). Suppose X is a metric space and K ⊆ X is compact.Then for any sequence
{xn } ⊆ K, there exists a subsequence {xnk }, such that xnk → x ∈ X. Then K is closed.
Assume K is unbounded, then for every finite subset {x1 , · · · , xn } ⊂ K, there exists
xn ∈ K, such that d(xn , x0 ) > n for n = 1, 2, · · · , and consequently we can find an infinite
sequence {xn }∞
n=1 , such that d(xn , xm ) > M for all m ̸= n. Obviously this sequence does
not contain a convergent subsequence, thus K is not compact.
(ii). Suppose W is a closed subset of K compact space. Let U be an open cover of W ,
since K/W is open then U ∪ (K/W ) is an open cover of K. Because K is compact, there
is a finite open subcover of K and these subcovers (exclude the set K/W ) cover W .
2. Let (X, d) be a complete metric space, and Y ⊂ X. Prove that (Y, d) is complete if and
only if Y is a closed subset of X.
Proof. ” ⇒ ”: (Y, d) is complete, then for any convergent sequence of Y , it is a Cauchy
sequence, then it converges in Y . Thus Y is closed.
” ⇐ ”: Now Y is a closed subset of X, then for any Cauchy sequence {xn } → x ∈ X, we
can see that x ∈ Y . This means that Y is complete.
3. Let (X, d) be a metric space, and let {xn } be a sequence in X. Prove that if {xn } has
a Cauchy subsequence, then, for any decreasing sequence of positive ϵk → 0, there is a
subsequence {xnk } of {xn } such that
d(xnk , xnl ) ≤ ϵk
for all k ≤ l.
Proof. Given a Cauchy subsequence of {xn }, still denoted by {xn }.For any ε1 > 0, there
exists N1 such that for all l ≥ k ≥ N1 , d(xk , xl ) ≤ ε1 , denote N1 = n1 . Now, for any
ε2 > 0, there exists N2 > N1 such that for all l ≥ k ≥ N2 , d(xk , xl ) ≤ ε2 , denote
N2 = n2 . Continue this process and we can get a subsequence {xnk } of {xn } satisfying
d(xnk , xnl ) ≤ εk , for all k ≤ l.
4. Suppose that f : X → R is lower semicontinuous and M is a real number. Define fM :
X → R by
fM (x) = min(f (x), M ).
Prove that fM is lower semicontinous.
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Proof.
lim inf fM (xn ) = lim inf min(f (xn ), M )
n→∞
n→∞
≥ min(lim inf f (xn ), lim inf M )
n→∞
n→∞
≥ min(f (x), M )
= fM (x)
5. Let f : X → R be a real-valued function on a set X. The epigraph epif of f is the subset
of X × R consisting of points that lie above the graph of f :
epif = {(x, t) ∈ X × R|t ≥ f (x)}.
Prove that a function is lower semicontinuous if and only if its epigraph is a closed set.
Proof. ” ⇒ ”: Take{(xn , tn )} ⊂ epif and (xn , tn ) → (x, t), f (xn ) ≤ tn . Pass limit inferior
on both sides and we have
t = lim inf tn ≥ lim inf f (xn ) ≥ f (x)
n→∞
n→∞
So (x, t) ∈ epif , which means epif is closed.
” ⇐ ”: Suppose epif is closed. Let lim inf f (xn ) = A, then there exists a subsequence xnk
n→∞
such that lim f (xnk ) = A. So (xnk , f (xnk )) → (x, A) ∈ epif .
n→∞
⇒ lim inf f (xn ) = A ≥ f (x)
n→∞
This shows that f (x) is lower semicontinuous.
6. Suppose that {xn } is a sequence in a compact metric space with the property that every
convergent subsequence has the same limit x. Prove that xn → x as n → ∞.
Proof. Suppose otherwise, then xn does not converge to x, so there exists a neighborhood
U of x such that infinitely many xk * U . And we can pick a subsequence from them,
denoted by yk = xnk , yk * U for ∀k ∈ N . Since the metric space is compact, then {yk }
has a convergent subsequence {ynk } and of course {ynk } is the subsequence of {xn }. But
since ynk * I, so ynk 9 x, this is a contradiction.
7. Prove theorem 1 on page 3.
Proof. (i).Obvious.
∑
∑
∑
(ii).∀x, y ∈ Yα , we can write x = xi , y = yi for xi , yi ∈ Yα . Then for ∀a, b ∈ R, we
α
i
have
ax + by =
∑
i
axi +
i
∑
byi =
i
∑
i
2
(axi + byi ) ∈
∑
α
Yα
(iii).∵ Yα ⊂ X, ∴ ∩Yα ⊂ X. Then ∀x, y ∈ ∩Yα , we have ax + by ∈ Yα for all α, thus ∩Yα
α
α
α
is a linear subspace of X.
∞
∞
(iv).Yn ⊂ Yn+1 , then ∪ Yn ⊂ X. And for ∀x, y ∈ ∪ Yn , there exist Yk , Yl , k ≤ l such that
n=1
n=1
∞
x ∈ Yk , y ∈ Yl . By the nondecreasing property of Y we know that ax+by ∈ Yl ⊂ ∪ Yn .
n=1
8. Prove theorem 3 on page 5.
Proof. (i).∀u1 , u2 ∈ M Y , there exist y1 , y2 ∈ Y such that M y1 = u1 and M y2 = u2 . Then
u1 + u2 = M y1 + M y2 = M (y1 + y2 ) ∈ M Y , and ku1 = kM y1 = M (ky1 ) ∈ M Y .So M Y
is a linear subspace of U .
(ii).∀x1 , x2 ∈ M −1 V , there exist v1 , v2 ∈ V , such that M −1 v1 = x1 and M −1 v2 = x2 .
Then x1 + x2 = M −1 v1 + M −1 v2 = M −1 (v1 + v2 ) ∈ M −1 V , kx1 = kM −1 v1 = M −1 (kv1 ) ∈
M −1 V . So M −1 V is a linear subspace of X.
9. Prove theorem 4 on page 5.
Proof. We use mathematical induction method:
For n = 1 and n = 2, they are obviously correct.
Suppose now n = k is true, then we consider the case n = k + 1. We divide it into two
situations:
(1) ak+1 = 1,a1 = a2 = · · · = ak = 0. Then it’s true since x = xk=1 ∈ K and satisfies the
conditions.
(2) 0 ≤ ak+1 < 1. Then we have the following calculation:
k+1
∑
an xn = (1 − ak+1 )
n=1
k
∑
n=1
an
xn + ak+1 xk+1
1 − ak+1
provided
k
∑
n=1
an
=1
1 − ak+1
By the case n = k we know that
k
∑
n=1
⇒
an
xn ∈ K
1 − ak+1
k+1
∑
an xn ∈ K
n=1
This finished our proof.
10. Prove theorem 6 on page 6.
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Proof. (i). Since Sˆ contains S and belongs to every convex sets containing S, then Sˆ is
the smallest.
(ii). Since convex hull contains S and is convex, then all of the convex combinations belong
to convex hull. On the other hand, all the combinations form a convex set containing S,
then Sˆ belongs to the set.
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