MATH 403 ANALYSIS I - SPRING 2010
SOLUTIONS to HOMEWORK 5
Problem 1. Let f : (X, d) → (Y, ρ). Show that the following are equaivalent:
(a) f is continuous.
(b) f (A) ⊂ f (A) for every A ⊂ X.
(c) f −1 (B ◦ ) ⊂ (f −1 (B))◦ for every B ⊂ Y .
Solution: (a) =⇒ (b) Assume that f is continuous. Let A ⊂ Y . Then f (A) is
closed in Y since f is continuous, f −1 (f (A)) is closed in X. Since A ⊂ f −1 (f (A)),
it follows that A ⊂ f −1 (f (A)) since A is the smallest closed set containing A. Hence
f (A) ⊂ f (A), as claimed.
(b) =⇒ (c) Assume that f (A) ⊂ f (A) for every A ⊂ X. Take any subset B of
Y and let x ∈ f −1 (B ◦ ). Then f (x) ∈ B ◦ . We claim that x ∈ f −1 (B)◦ . If this is
not the case, then x is an adherent point of (f −1 (B))c . That is, x ∈ (f −1 (B))c .
By assumption, f ((f −1 (B))c ) ⊂ (f −1 (B))c ⊂ B c . Thus, f (x) ∈ B c . This means
that any ball around f (x) intersects B c contradicting that f (x) ∈ B ◦ . Hence x ∈
(f −1 (B))◦ and f −1 (B ◦ ) ⊂ (f −1 (B))◦ as claimed.
(c) =⇒ (a) Take an open subset U of Y . We have to show that f −1 (U ) is open
in X. Since U = U ◦ and by assumption f −1 (U ◦ ) ⊂ (f −1 (U ))◦ , it follows that
f −1 (U ) ⊂ (f −1 (U ))◦ . Since (f −1 (U ))◦ ⊂ (f −1 (U )), we have f −1 (U ) = (f −1 (U ))◦
showing that f −1 (U ) is open in X.
Problem 2.
(a) Let {xn } and {yn } be sequences in a metric space (X, d). Assume that {xn }
is Cauchy and that d(xn , yn ) → 0. Show that {yn } is Cauchy in (X, d).
(b) A subset D of X is called dense if D = X. Suppose that every Cauchy
sequence of points in D converges to some point of X. Show that (X, d) is
complete.
Solution: (a) Take ε > 0. Then there is N1 such that
for all n, m ≥ N1 .
d(xn , xm ) < ε/3,
Moreover, there is N2 such that
for all n ≥ N2 .
d(xn , yn ) < ε/3
Set N := max{N1 , N2 } and let n, m ≥ N . Then using the triangle inequality
d(yn , ym ) ≤ d(yn , xn ) + d(xn , xm ) + d(xm , ym ) < ε/3 + ε/3 + ε/3 = ε
showing that (yn ) is Cauchy.
(b) Let (xn ) be Cauchy in (X, d). Since D = X, xn ∈ D. This implies, in
particular, that B1/n (xn ) ∩ D 6= ∅. Hence for every n there is yn ∈ D such that
d(xn , yn ) < 1/n. This means that d(xn , yn ) → 0. By (a), the sequence (yn ) is
Cauchy. By assumption there is x ∈ X such that d(yn , x) → 0. Then
d(xn , x) ≤ d(xn , yn ) + d(yn , x) → 0
showing that (X, d) is complete.
Problem 3.
1
2
(a) Consider R with the metric
ρ(x, y) = |arctan x − arctan y| .
Is (R, τ ) a complete metric space?
(b) Consider R with the metric
τ (x, y) = x3 − y 3 ,
x, y ∈ R.
Is (R, τ ) a complete metric space?
Solution: Take a sequence xn = n. Since arctan n → π/2, for given ε > 0, there
is N so that |arctan n − π/2| < ε/2 for n ≥ N . Hence
ρ(xn , xm ) = |arctan n − arctan m| ≤ |arctan n − π/2| + |arctan m − π/2| < ε
for n, m ≥ N . So, (xn ) is Cauchy with respect to τ . Assume that ρ(xn , x) →
0 for some x ∈ R. Then π/2 = arctan x which is impossible since arctan(R(⊂
(−π/2, π/2).
(b) Let (xn ) be Cauchy with respect to τ . Then (x3n ) is Cauchy with respect to the
standard metric
in R. Since R with the standard metric is complete,
isy ∈ R
there
√
such that x3n − y → 0. Set x = 3 y. Then τ (xn , x) = x3n − x3 = x3n − y → 0.
Consequently, (R, τ ) is complete.
Problem 4.
(a) P
Let (X, k·k) be a normed space and let (xn ) be a sequence in X satisfying
∞
n=1 kxn k < ∞. Show that (xn ) is a Cauchy sequence.
(b) Let (X, k·k) be a normed space. Show that (X, k·k) is complete if and only
if B 1 (0) = {x ∈ X| kxk ≤ 1} is complete.
(c) Let S be the vector space of all real sequences x = (xn ) such that xn = 0
for all but finitely many n. Show that S is not complete with respect to
the norm kxk = supn |xn |.
Pm
P∞
Solution: (a) Let ε > 0. Since n=1 kxn k < ∞, there is N such that k=n kxk k <
ε for all m > n ≥ N. So,
kxn − xm k ≤ kxn k + kxm k ≤
m
X
k=n
kxk k < ε
for all m > n ≥ N , showing that (xn ) is Cauchy in (X, k·k).
(b) A closed subset of a complete metric space is complete. Consequently, if (X, k·k)
is complete, the closed ball B 1 (0) is complete. Conversely, assume that B 1 (0) is
complete and let (xn ) be a Cauchy sequence in (X, k·k). Then (xn ) is bounded.
Hence there is R > 0 such that xn ∈ BR (0). Then yn = xn /R ∈ B1 (0). The
sequence (yn ) is Cauchy in B 1 (0) and so it converges to y ∈ B 1 (0). Setting x = Ry,
we have kxn − xk = kRyn − Ryk = R kyn − yk → 0. Hence (X, k·k) is complete as
claimed.
(c) For every n, let Xn = (1, 1/2, 1/3, . . . , 1/n, 0, 0, . . .). Then Xn ∈ S. The
sequence (Xn ) is Cauchy. Indeed, take ε > 0 and let N be such that 1/N < ε.
Then for m > n ≥ N ,
kXn − Xm k =
1
1
<
< ε.
n+1
N
3
Assume that (Xn ) converges to X ∈ S. Then X = (x1 , x2 , . . .) and by definition of
S there is N such that xn = 0 for all n ≥ N . Then for every n > N ,
1
kXn − Xk ≥
N
which implies that (Xn ) cannot converge to X.