Homework 6 Solutions
Math 501
Due October 10, 2014
Exercise 1
Let M be a metric space and let N = {0, 1} be a metric space by giving it
the metric inherited from R. Prove that M is connected if and only if every
continuous function f : M → N is constant.
Suppose M is connected. Let f : M → N be continuous. It is not difficult
to see that the sets A = {0} and B = {1} are open in N . Therefore f −1 (A)
and f −1 (B) are open in M . Moreover
f −1 (A) ∩ f −1 (B) = f −1 (A ∩ B) = ∅,
f −1 (A) ∪ f −1 (B) = f −1 (A ∪ B) = M.
Therefore since M is connected, it follows that either f −1 (A) = ∅ or f −1 (B) = ∅.
In either case, it follows that f is constant.
Suppose M is disconnected. Let A be a proper clopen subset of M . Let
f : M → N be the function given by
1 if x ∈ A
f (x) =
0 if x 6∈ A.
Note that
f −1 (N ) = M,
f −1 ({0}) = Ac ,
f −1 ({1}) = A,
f −1 (∅) = ∅.
It follows that the pre-image of every open subset of N is open, which implies
f is continuous. Moreover, f is not constant.
Exercise 2
(a) Let N be closed in M and let K ⊂ N .
Suppose K is closed in N . Then there exists L ⊂ M closed in M such that
K = L ∩ N . Therefore since L and N are closed in M , and because the
intersection of closed sets is closed, we have that K is closed in M .
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Conversely suppose that K is closed in M . Let L = K. Then K = L ∩ N
is the intersection of N with a closed subset of M .
(b) Let N be open in M and let U ⊂ N .
Suppose U is open in N . Then there exists V ⊂ M open in M such that
U = V ∩ N . Therefore since L and N are open in M , and because the finite
intersection of open sets is open, we have that U is open in M .
Conversely suppose that U is open in M . Let V = U . Then U = V ∩ N is
the intersection of N with an open subset of M .
Exercise 3
Let f : M → R be a function. The graph of f is the set Γ = {(p, f (p)) ∈
M × R : p ∈ M }.
(a) Suppose f is continuous. Let gn = (pn , f (pn )) ⊂ Γ be a sequence that
converges to some (p, y) ∈ M × R. To show that Γ is closed, it suffices to
show that (p, y) ∈ Γ. To this end, it suffices to show that y = f (p).
Let > 0. Observe that since gn → (p, y), we have pn → p and f (pn ) → y.
Moreover, because f is continuous f (pn ) → f (p). Thus there exist N1 , N2
such that
|f (pn ) − y| < /2, |f (pm ) − f (p)| < /2
for all n ≥ N1 , m ≥ N2 . Therefore, if N = max{N1 , N2 }, then
|f (p) − y| ≤ |f (p) − f (pn )| + |f (pn ) − y| < .
The result now follows.
(b) Observe that the set H = {(x, y) ∈ R2 : xy = 1, x, y > 0} is the graph of the
function f : (0, ∞) → R given by f (x) = 1/x which is continuous. Part (a)
implies that H is closed in (0, ∞) × R. Although it might seem promising
to use this fact to show that H is closed in R2 , it is easier to prove this
directly.
Let (xn , yn ) ⊂ H be a sequence that converges to (x, y) ∈ R2 . To show that
H is closed, it suffices to show that (x, y) ∈ H. Indeed, observe that xn → x
and yn → y, which implies that xn yn → xy. Moreover, since xn yn = 1 for
all n ≥ 1, it follows that xy = 1, which implies (x, y) ∈ H.
(c) Let X = {(x, 0) ∈ R2 : x ∈ R}. It was shown in Exercise 7 of Homework 4
that X is closed in R2 . It follows that X ∪ H is the finite union of closed
sets in R2 and hence closed in R2 .
(d) Let S = H ∪ X. Note that by part (c), S is closed in R2 . It follows from
Exercise 2 that both H and X are closed in S since they are closed in R2 .
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