Thread: topology/metric spaces questions

Hi everyone, I'm studying for a topology exam. Any help would be really appreciated.

1. (a) Let X be the interval (0, 1/3) with usual Euclidean metric. Show that f : X — » X defined by f(x) = x^2
is a contraction, but f does not have a fixed point in X. Why does this not contradict the Banach
fixed point theorem?
(b) Let (X, d) be a complete metric space and f : X — » X. Define g(x) =f(f(x)), that is, g = f o f. Assume that the map g : X — > X is a contraction. Prove that f has a unique fixed point.

2. (a) Let (X, d) be a metric space and let {fn} be a sequence of continuous functions, fn : X — > R, for n € N. Prove that if {fn} converges uniformly to f : X — > R, then f is a continuous function.
(b) Let fn(x) = (1-xⁿ)/(1+xⁿ) for x € [0, 1] and n € N. Find the pointwise limit f of the sequence {fn}, and determine whether the sequence is uniformly convergent on the interval [0, 1].


3. (a) Let X be a topological space, and let A and B be compact subsets of X. Prove that the union A U B is a compact subset of X.
(b) Let f : X —> Y be a continuous map between topological spaces. Prove that if X is compact, then f(X) is compact.

Originally Posted by laura_d

Hi everyone, I'm studying for a topology exam. Any help would be really appreciated.

1. (a) Let X be the interval (0, 1/3) with usual Euclidean metric. Show that f : X — » X defined by f(x) = x^2
is a contraction, but f does not have a fixed point in X. Why does this not contradict the Banach
fixed point theorem?

on we have , so:



So is a contraction on .

There is no fixed point as that would require that:



have a root in and it does not.

This does not contradict the fixed point theorem as is not closed, complete or whatever the equivalent condition in the version of the theorem you are using is.

CB