# fixed point of homeomorphism and compactness of a complete metric space.

I need to know that the following statements if true or false:

1. Every homeomorphism of $S^2\rightarrow S^2$ has a fixed point.

2. Let $X$ be a complete metric space such that distance between any two point is less than $1$, Then $X$ is compact.

well, for 1 I see that it is false as $S^2$ is not convex so Brauer Fixed point Theorem can not be applied?

for 2 I thought that it will be compact, if not then my intuition says that it will be not sequentially compact or violates the definition of compactness?

Thank you for the help

1. $x \mapsto -x$.
2. $X = \mathbb{R}$, $d(x,y) = \tfrac{1}{2}$ if $x\neq y$ and $d(x,x) = 0$.
1. While the fact that $S^2$ is not convex means that Brouwer's fixed point theorem cannot be applied, it doesn't mean that it doesn't have the fixed point property. Any non-convex set which is homeomorphic to a convex compact subset of $\mathbb{R}^n$ will be an example. However, the antipodal map $x \mapsto -x$ is an example of a homeomorphism of $S^2$ which has no fixed points.
2. Take the closed unit ball of radius $\frac{1}{4}$ of any infinite dimensional Banach space.