Fixed points sets sine of the topologist Consider the sine of the topologist set
$X:=\{(x,\sin(1/x)):x\in (0,1]\}\cup \{(0,y):y\in [-1,1]\}$,
in the euclidean space $\mathbb{R}^{2}$. It is not very hard to show, see this proof, that every continuous $f:X\longrightarrow X$ has some fixed point.
My question is the following: Can we find a closed subset of $X$, say $C$, such that $C$ is not the fixed point set of any continuous $f:X\longrightarrow X$? Intuitively, it i seems that such set $C$ could be of the form $C_{1}\cup C_{2}$ with $C_{1}$ and $C_{2}$ closed sets of the arc-wise connected component of $X$, what do you think?
Many thanks in advances for your comments.
 A: The answer is ''yes''. Let $L =  \{(0,y):y\in [-1,1]\}$, $R = \{(x,\sin(1/x)):x\in (0,1]\}$. Define
$$C = X \cap ([0,1] \times [-1,0] .$$
Assume $C$ is the fixed point set of a continuous $f : X \to X$. Since $f(R)$ is path-connected and $C \cap R = f(C \cap R) \subset f(R)$, we must have $f(R) = R$. Hence $X = \overline{R} = \overline{f(R)} \subset \overline{f(X)} = f(X)$ because $f(X)$ is compact. This means that $f$ is surjective and we conclude $f(L) = L$. We therefore have $f(0,y) = (0,F(y))$ with a continuous surjection $F : [-1,1] \to [-1,1]$. Hence there exists $a \in [-1,1]$ such that $F(a) = 1$. Set $b = F(1)$. By construction $[-1,0]$ is the fixed point set of $F$ and we conclude $a \in (0,1), b \in [-1,1)$.
Define $u : [a,1] \to [-1,1]^2, u(y) = (y,F(y))$. Let $\Delta$ denote the diagonal of $[-1,1]^2$. We know that $u(y) \notin \Delta$ for $y \in [a,1]$ because $F$ does not a fixed point in this interval. Hence $u$ is a path in $A = [-1,1]^2 \backslash \Delta$. The space $A$ has two components $A^+, A^-$, where $A^+$ lies above the diagonal. But $u(a) = (a,1) \in A^+, u(1) = (1,b) \in A^-$ which is impossible.
Therefore $C$ is not the fixed point set of any $f$.
