Show that any non-empty, non-compact convex subset of $\mathbb{R}^n$ lacks the fixed point property 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. 
 A: 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 
