# Thread: Metric Spaces/Topology help

1. ## Metric Spaces/Topology help

Any help with these would be hugely appreciated

(1) Let (X, d) and (Y, d'
) be metric spaces, and let f : X — » Y be continuous with f(X) = Y. Show that if (X,d) is complete and
d(x,y) < kd'(f(x),f(y)) for some constant k and all x,y € X, then (Y, d') is complete.

(2) Let A be a non-empty compact subset of X. Prove that there exist points a,b € A such that d(a, b) = sup{d(x,y) : x,y € A}.
(3)(a) Prove that a topological space X is connected if and only if every continuous map f : X —» {0,1} is constant, where {0,1} is equipped with the discrete topology.
(b) Let S
1 be the unit circle {(x,y) € R2 : x2 + y2 = 1} in R2 with the topology induced by the standard topology of R2 . Assume that
f : S1 —» R is a continuous map. Prove that there exists a point
z = (x,y) € S1 such that f(z) = f(-z). (Hint: Consider the function
g(z) = f(z) - f(-z)).

2. Originally Posted by laura_d
Any help with these would be hugely appreciated

(2) Let A be a non-empty compact subset of X. Prove that there exist points a,b € A such that d(a, b) = sup{d(x,y) : x,y € A}.

Hint: $\displaystyle d(a,b)=\text{diam}\left(A\right)$, consider some Cauchy Sequence $\displaystyle \left\{p_n\right\}$ on $\displaystyle A$, now use the fact that if $\displaystyle \left\{p_n\right\}$ is a Cauchy Sequence that $\displaystyle \text{diam}\left(A\right)$ exists

3. Originally Posted by laura_d
Any help with these would be hugely appreciated

(1) Let (X, d) and (Y, d') be metric spaces, and let f : X — » Y be continuous with f(X) = Y. Show that if (X,d) is complete and
d(x,y) < kd'(f(x),f(y)) for some constant k and all x,y € X, then (Y, d') is complete.

(2) Let A be a non-empty compact subset of X. Prove that there exist points a,b € A such that d(a, b) = sup{d(x,y) : x,y € A}.
(1) You need to show that every Cauchy sequence in Y converges. You are told that f is surjective, so a Cauchy sequence in Y must be of the form $\displaystyle \{f(x_n)\}$ for some sequence $\displaystyle \{x_n\}$ in X. Use the information in the question to deduce that $\displaystyle \{x_n\}$ converges to some point $\displaystyle z\in X$, and conclude that $\displaystyle \{f(x_n)\}$ converges to f(z).

(2) Let $\displaystyle D = \sup\{d(x,y):x,y\in A\}$. By the definition of supremum, there are sequences $\displaystyle \{x_n\},\ \{y_n\}$ in A such that $\displaystyle d(x_n,y_n)\to D$ as $\displaystyle n\to\infty$. Since A is compact, $\displaystyle \{x_n\}$ has a convergent subsequence. Replacing $\displaystyle \{x_n\}$ by this subsequence, you can assume that $\displaystyle \{x_n\}$ converges, to a say. Then do the same for the other sequence $\displaystyle \{y_n\}$, so that this also converges, to b say. Then conclude that $\displaystyle d(x_n,y_n)\to d(a,b)$.