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Homethe topologist's sine curve has the fixed point property

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# the topologist’s sine curve has the fixed point property

The typical example of a connected space that is not path connected (the topologist’s sine curve) has the fixed point property.

Let $X_{1}=\{0\}\times[-1,1]$ and $X_{2}=\{(x,\sin(1/x)):0<x\leq 1\}$, and $X=X_{1}\cup X_{2}$.

If $f:X\to X$ is a continuous map, then since $X_{1}$ and $X_{2}$ are both path connected, the image of each one of them must be entirely contained in another of them.

If $f(X_{1})\subset X_{1}$, then $f$ has a fixed point because the interval has the fixed point property. If $f(X_{2})\subset X_{1}$, then $f(X)=f(cl(X_{2}))\subset cl(f(X_{2}))\subset X_{1}$, and in particular $f(X_{1})\subset X_{1}$and again $f$ has a fixed point.

So the only case that remains is that $f(X)\subset X_{2}$. And since $X$ is compact, its projection to the first coordinate is also compact so that it must be an interval $[a,b]$ with $a>0$. Thus $f(X)$ is contained in $S=\{(x,\sin(1/x)):x\in[a,b]\}$. But $S$ is homeomorphic to a closed interval, so that it has the fixed point property, and the restriction of $f$ to $S$ is a continuous map $S\to S$, so that it has a fixed point.

This proof is due to Koro.

## Mathematics Subject Classification

55M20*no label found*54H25

*no label found*47H10

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