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The Fixed-Point Property is stated as such: Every continuous self-map admits a fixed point.

Attempt: Since compact sets in $\mathbb{R}^n$ are necessarily closed and bounded, we consider two cases. Where the subset is non-closed and when the set is non-bounded. For a non-closed set, if we can find a mapping such that the only fixed points lie on the boundary, then we will have what we want, although we also need this to be a continuous mapping. I do not have a clue what to do with the unbounded case. Please help.

Mathemphetamine
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    The idea with bounded and unbounded case is the same (in fact, the "infinity" plays the same role as the point missing from the boundary). Your map needs to push all points closer to the "missing" boundary point (or infinity). – xyzzyz May 19 '16 at 22:51
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    Hint: Show that each unbounded convex subset $C$ of $R^n$ contains a ray $R$. Can you construct a continuous map $C\to R$ which does not have a fixed point? Now, do the same for bounded non-closed convex subsets, but use half-open intervals instead of rays. – Moishe Kohan May 19 '16 at 23:06
  • Additional Q: What can we say about a non-compact non-convex subset? – DanielWainfleet May 20 '16 at 02:55
  • @user254665: If the subset is noncompact and contains a properly embedded copy of $[0,1)$, the same method as in the convex case yields self-maps without fixed points. I am not sure otherwise. – Moishe Kohan May 20 '16 at 18:14
  • @studiosus. I don't know a general answer. The map $f(x)=1-x$ maps the Cantor set $C$ homeomorhically to itself with no fixed point . And $g(x)=x+1$ maps $\cup_{n\in Z}(n+C)$ to itself with no fixed point. – DanielWainfleet May 21 '16 at 22:35
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    @user254665: I think you should ask a separate question, since this has very little to do with the question asked in this post. As for interesting (noncompact) examples, I think the union of the graph of $\sin(1/x)$ and the origin in the plane has fixed-point property. – Moishe Kohan May 22 '16 at 16:29

1 Answers1

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Theorem 1. Suppose that $X$ is a subset of ${\mathbb R}^n$ which contains a closed subset $C\subset X$ such that $X$ is either a ray or a half-open line segment $[AB)$. Then $X$ does not satisfy the fixed point property, i.e. there exists a continuous self-map $f: X\to X$ without fixed points.

Proof. Consider the nearest-point projection $p: X\to C$: Since $C$ is closed in $X$, for every $x\in X$ the infimum $$ \inf\{d(x,c): c\in C\} $$ is attained at some $c\in C$. Since $C$ is convex, the point $c:=p(x)$ is unique and the function $x\mapsto p(x)$ is continuous (actually, 1-Lipschitz). The map $p$, of course, is the identity on $C$. Now,let $C$ be parameterized as $t\mapsto \rho(t), t\in [0,\infty)$, consider the continuous map $$ h: C\to C, h(\rho(t))= \rho(t+1). $$ Lastly, let $f:= h\circ p$. This map cannot have a fixed point in $X$ since $f(X)\subset C$ and $f$ has no fixed points in $C$ either. qed

Corollary. If $X\subset {\mathbb R}^n$ is a noncompact convex subset, then $X$ does not satisfy the fixed point property.

Proof. Pick a point $x_0\in X$. If $X$ is bounded but not closed, there is a bounded sequence $x_i\in X$ converging to a boundary point $y\in cl(X) - X$. Considering the limit $x_0y$ of the sequence of line segments $x_0x_n$ and using convexity of $X$, we conclude the halfopen segment $C=[x_0y)$ is contained in $X$. By the construction, $C$ is closed in $X$.

If $X$ is unbounded, consider a sequence $(x_i)$ in $X$ such that $|x_i|\to \infty$. A subsequence, in $x_0x_i$ converges to a ray $C$ contained in $X$.

Now, applying Theorem 1, we conclude that $X$ does not satisfy the fixed point property. qed

With a bit more work, one can similarly prove:

Theorem 2. Suppose that $X$ a CW complex for which there exists a 1-1 continuous map $r: [0,\infty)\to X$ whose image $C$ is closed. Then $X$ also does not have the fixed point property.

Example. Let $X\subset {\mathbb R}^2$ be the subset equal to the union of the graph $G$ of the function $y=\sin(1/x), x\in (0,1]$ and the origin $\{(0,0)\}$ in ${\mathbb R}^2$. Then $X$ is (clearly) noncompact but satisfies the fixed point property.

Proof. Indeed, let $f: X\to X$ be a continuous function. Then $f(G)\cap \{(0,0)\}= \emptyset$. If $f((0,0))=(0,0)$, we are done. Otherwise, $f: G\to G$ and $f((0,0))= (x_0,\sin(1/x_0))$ for some $x_0\in (0,1]$. Conjugating with the homeomorphism $h: x\mapsto (x, \sin(1/x))$ we convert $f$ to a continuous function $$ F= h^{-1}\circ f\circ h: (0,1]\to (0,1]. $$ It is clear that $f$ has a fixed point iff $F$ does. If $F(1)=1$, we are again done. Assume, therefore, that $F(1)<1$. Continuity of $f$ implies that $$ \lim_{n\to \infty} F(\frac{1}{\pi n}) = x_0>0. $$ It follows (from $F(1)<1$ and the above limit) that the graph of $F$ contains points both above and below the line $y=x$. Hence, $F$ has a fixed point in $(0,1]$. qed

Moishe Kohan
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