MATH 3610.005 HOMEWORK 1
(1) Proof. Given x ∈ (a, b), let (xn ) be a sequence of rationals in (a, b) converging to x. Since f is continuous at x, f (x) = lim f (xn ) = limn→∞ 0 = 0. (2) Proof. Suppose f is continous on R and let H be a closed set. Then R \ H
is open by Theorem 13.7. By Corollary 21.15, f −1 (R \ H) is open. But
f −1 (R \ H) = R \ f −1 (H) by Theorem 7.14(g). Thus R \ f −1 (H) is open
and f −1 (H) is closed by Theorem 13.7. The converse is similar.
(3) Proof. Consider the function g : [a, b] → [a, b] definded as g(x) = f (x) − x.
First note g is continuous and
g(a) = f (a) − a ≥ a − a = 0
g(b) = g(b) − b ≤ b − b = 0
Also note that if g(a) = 0 or g(b) = 0, we will have f (a) = a or f (b) = b
and we are done. Without loss of generality let g(a) < 0 < g(b). Then as
g is continuous by lemma 22.5 ∃c ∈ (a, b) such that g(c) = 0. But
g(c) = 0 ⇐⇒ f (c) − c = 0 ⇐⇒ f (c) = c
(4) Proof. Theorem 22.10 implies that f ([a, b]) is a compact interval in R. If
f ([a, b]) contained more than one point, it would have to contain irrational
points, a contradiction. Thus f ([a, b]) is a single point and f is constant on
[a, b].
(5) (a)

if x ∈ [ 12 , 1]
 x − 21
1
if x = n1 and n = 3, 4, 5, . . . ,
f (x) =
 n−1
x
otherwise
Then f is bijective from [1/2, 1] onton [0, 1/2] adn f is bijective from
[0, 1/2) onto [0, 1/2].
(b) Proof. Suppose f were continuous. Then f atains its maximum at
two distinct points, say M1 and M2 . Likewise, f attains it’s minimum
at two distinct points, say m1 and m2 . Considerting the possible
orderings of the m0 s and the M 0 s it is evident there exists some value
that is attained 3 times. Since for each y ∈ R, f −1 ({y}) is either empty
or contains exactly two points, no such fucntion can be continuous. (6) Proof. Let A, B ⊆ X be compact. Then consider (A ∩ B). As A and B
are both compact A and B are bounded. Let ma ,mb ,Ma and,Mb be lower
bounds for A and B respectively. As (A ∩ B) ⊆ A and (A ∩ B) ⊆ B,
let m = max{ma , mb } and M = min{Ma , Mb }. Thus (A ∩ B) is bounded
bellow by m and above by M . Since A and B are closed (X \A)∪(X \B) =
X \ (A ∩ B) is open. Thus (A ∩ B) is closed. So (A ∩ B) is closed and
bounded, thus compact.
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