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I'm currently doing a course on the theory of metric spaces.

This is the version of Schauder fixed point theorem from my course:

Let $(X,\|\cdot\|)$ be Banach and $C\subset X$ a closed, bounded, convex subset. If $ f\colon C \to C $ is a compact map, then there is a $x_*\in C$ such that $f(x_*)=x_*$.

Recall:

A continuous map $f\colon C\to C$ (where $C\subset X$) is compact if for any bounded sequence $(c_n)$ in $C$ the sequence $\{f(c_n)\}$ has a convergent sub-sequence.

This is my question:

If I do not demand that $f$ is continuous, (i.e. I call a map $f\colon C\to C$ compact if for any bounded sequence $(c_n)$ in $C$ the sequence $\{f(c_n)\}$ has a convergent sub-sequence), does (my version of) the Schauder fixed point theorem still hold?

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1 Answer 1

No. Let $x_1, x_2 \in C$, $x_1 \ne x_2$. Define $f\colon C \to C$ by \[ f(x) = \begin{cases} x_1 & x \ne x_1\\ x_2 & x = x_1 \end{cases} \] Then $f$ has a compact image, but no fixed point.

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Thank you for helping me :D –  Jim Jul 6 '12 at 19:59

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