Download MATH 429 - Fall 2005 Problem 8, Page 111. Let X be a topological

MATH 429 - Fall 2005
Problem 8, Page 111. Let X be a topological space. Let Y be an ordered
set with its order topology. let f, g : X → Y be continuous.
(a) Show that the set A = {x | f (x) ≤ g(x)} is closed in X.
Solution:
Case 1: If f = g on X then A = X is closed.
Case 2: Suppose f 6= g. The complement of A in X is
X − A = {x | g(x) < f (x)}
We will show that X − A is open. Suppose that X − A is non-empty and
pick an arbitrary element x0 ∈ X − A. Let a, b and c be elements in Y such
that
a ≤ g(x0 ) < b ≤ f (x0 ) ≤ c
In the order topology, [a, b) and [b, c] are open sets. Because f and g are
continuous, g −1 ([a, b)) and f −1 ([b, c]) are open in X. Their intersection is an
open neighborhood of x0 which is entirely contained in X − A. Since x0 is
an arbitrary element of X − A, we conclude that X − A is open.
(b) Show that the function h : X → Y defined by h(x) = min{(f (x), g(x)}
is continuous.
Solution: Let B = {x | g(x) ≤ f (x)}. As above, one can show that B is
closed in X. We have X = A ∪ B. Moreover, h(x) = f (x) for every x ∈ A
and h(x) = g(x) for every x ∈ B. By the pasting lemma, h is continuous.
Problem 2, Page 152. Let {An } be a sequence of connected subspaces of
a topological space X such that An ∩ An+1 6= ∅ for all n. Show that ∪An is
connected.
Solution: We prove by induction, that ∪n1 Ak is connected for all n.
First, A1 ∪ A2 is connected because Theorem 23.3 states that the union of
a collection of connected subspaces of X that have a point in common is
connected. Now, suppose that Un = ∪nk=1 Ak is connected for some n ≥ 1.
Theorem 23.3 implies that An ∪ An+1 is connected. Applying Theorem 23.3
again, we conclude that (∪nk=1 Ak ) ∪ (An ∪ An+1 ) is connected. So, ∪n+1
k=1 Ak is
connected.
1
Problem 2, Page 158. Let f : S 1 → R be a continuous function on the
unit circle. Show that there a point p such that f (p) = f (−p).
Solution: Let g be the function defined by g(p) = f (p) − f (−p). Consider the points N = (1, 0) and S = −N = (−1, 0). One can check that
g(N ) = −g(S). If g(N ) > 0 then g(S) < 0 and vice versa. Applying the
Intermediate Value Theorem, we conclude that there is a point p such that
g(p) = 0, that is, f (p) = f (−p).
Note: The solution of Problem # 3, Page 158 is similar to the above one.
We apply the Intermediate Value Theorem using the function g(x) = f (x)−x
and the endpoints of [0, 1].
2