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A topological space $X$ has the fixed point property if every continuous map $f:X\to X$ has a fixed point (i.e., $x\in X$ such that $f(x)=x$).

Following wikipedia, I'll consider the following two variants, each a subspace of $\mathbb R^2$. Let $Y=\{(x, \sin\frac{1}{x}):x \in (0,1]\}$.

  1. The topologist's sine curve: $Y\cup\{(0,0)\}$.

  2. The closed topologist's sine curve: $Y\cup\{(0,y):y \in [-1,1]\}$, which is the closure of $Y$ in $\mathbb R^2$.

The "closed topologist's sine curve" has the fixed point property (proof below).

Question: Does the (not closed) "topologist's sine curve" have the fixed point property?

Proof that the closed topologist's sine curve has the fixed point property:

Let $X$ be the space and let $\pi:X\to[0,1]$ denote the projection onto the first coordinate. Suppose $f:X\to X$ is continuous. The path connected components of $X$ are $Y$ and the vertical interval $I=\{0\}\times[-1,1]$. A continuous function must map a path connected component into a path connected component.

Case 1: $f$ maps $I$ into $I$. Then $f$ has a fixed point in $I$ because a closed interval in $\mathbb R$ has the fixed point property.

Case 2: $f$ maps $I$ into $Y$. Since $I$ is compact, $f(I)$ is compact and $f(I)$ is contained in some open set of the form $V=\pi^{-1}((a,1])$ for some $a>0$. Then $f^{-1}(V)$ is an open nbhd of $I$. By the tube lemma, a nbhd of $I$ in $\mathbb R^2$ contains a set of the form $(-\varepsilon,\varepsilon)\times[-1,1]$. So in $X$, $f^{-1}(V)$ contains $\pi^{-1}([0,\varepsilon))$. In particular, we can infer that $Y$, which is a path connected component of $X$, must be mapped into $Y$. So the whole of $X$ is mapped into $Y$. Since $X$ is compact, $f(X)$ is a compact subset of $Y$, necessarily contained in $C=\pi^{-1}([b,1])$ for some $b>0$. $C$ is homeomorphic to the compact interval $[b,1]\subseteq\mathbb R$ and therefore has the fixed point property. Since $f$ maps $C$ into itself, it has a fixed point in $C$.

Simplified proof using ideas from Jakobian's answer:

(Sketch) Let $\pi:X\to[0,1]$ be the projection onto the first coordinate. Suppose $f$ has no fixed point in $Y$. Then the point $z_0=(1,\sin 1)$ satisfies $\pi(f(z_0))<\pi(z_0)$. By connectedness of $Y$, $\pi(f(z))<\pi(z)$ for all $z\in Y$. And by continuity of $f$ and the fact that $Y$ is dense in $X$, $\pi(f(z))\le\pi(z)$ for all $z\in X$. In particular, $f(I)\subseteq I$. So $f$ has a fixed point in $I$.

PatrickR
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1 Answers1

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Topologist's sine curve has fixed point property.

Let $X$ be topologist's sine curve as defined by Patrick, $\pi:X\to [0, 1]$ projection onto $x$-axis, $f:X\to X$ a continuous function without a fixed point, $X_1 = \{x\in X : \pi(x) < \pi(f(x))\}$ and $X_2 = \{x\in X : \pi(x) > \pi(f(x))\}$. Note that $\pi$ is a continuous bijection so that $X_1, X_2$ are open disjoint sets with $X = X_1\cup X_2$. Since necessarily $(0, 0)\in X_1$ and $(1, \sin 1)\in X_2$, they are non-empty. But this contradicts that $X$ is connected.

Jakobian
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