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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.

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  • $\begingroup$ It is not possible, because the identity map is a continuous function. $\endgroup$
    – Onil90
    May 23, 2018 at 8:54
  • $\begingroup$ @Onil90, the identity map (like every map) only has one fixed point set, namely $X$ itself; we are looking for full fixed point sets, not sets which are fixed by some map (possibly together with other points). $\endgroup$
    – Mees de Vries
    May 23, 2018 at 9:06
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    $\begingroup$ It seems to me that it is not possible. Consider the unit interval, and choose a closed sub-interval $A$. Is it not possible to construct a function whose restriction to $A$ is the identity, which is continuous, but has no other fixed points? You could then apply this to any closed subset of $X$, which is homeomorphic to a union of closed intervals. $\endgroup$
    – Damien
    May 23, 2018 at 12:04
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    $\begingroup$ Even $X$ itself is not homeomorphic to a union of closed intervals in any non-trivial sense; it is not homeomorphic to a disjoint union of closed intervals, or to a finite (possibly non-disjoint) union of closed intervals. The same is true for many of its subsets (even many of the subsets of the interval, actually). $\endgroup$
    – Mees de Vries
    May 23, 2018 at 12:35
  • $\begingroup$ It says in the proof you provided that $S = \{ (x, \sin(\frac{1}{x}) | x \in [a, b] \} $ is homeomorphic to a closed interval, for $a>0$. If $a=0$, $S$ would be homeomorphic to $[0, \infty)$, which is an infinite union of closed intervals. You could then choose the infinite union of intervals so that each one contains only one closed interval in $A$. $\endgroup$
    – Damien
    May 23, 2018 at 13:06

1 Answer 1

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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 $f$.

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  • $\begingroup$ Dear Paul Frost, Many thanks for your answer. $\endgroup$
    – user123043
    Sep 5, 2018 at 8:38

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