1. ## 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.

2. 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?
$d(f(x),f(y))=d(x^2,y^2)=\sqrt{x^4+y^4}$

on $(0,1/3)$ we have $x^4\le (1/9) x^2$, so:

$
d(f(x),f(y))=\sqrt{x^4+y^4} \le (1/3)\sqrt{x^2+y^2}=(1/3)d(x,y)
$

So $f$ is a contraction on $(0,1/3)$.

There is no fixed point as that would require that:

$x=x^2$

have a root in $(0,1/3)$ and it does not.

This does not contradict the fixed point theorem as $(0,1/3)$ is not closed, complete or whatever the equivalent condition in the version of the theorem you are using is.

CB