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 , 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.

Title	the topologist’s sine curve has the fixed point property
Canonical name	TheTopologistsSineCurveHasTheFixedPointProperty
Date of creation	2013-03-22 16:59:37
Last modified on	2013-03-22 16:59:37
Owner	Mathprof (13753)
Last modified by	Mathprof (13753)
Numerical id	6
Author	Mathprof (13753)
Entry type	Proof
Classification	msc 55M20
Classification	msc 54H25
Classification	msc 47H10