# 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

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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.