If $\omega:H \rightarrow H$ (into) is continuous and $H$ is compact, does $\omega$ fix any of its subsets other than the empty set?


the proof I learned from Henno Brandsma's answer: Let $X$ be compact Hausdorff (no metric is needed), and define $A_0 = X$, $A_{n+1} = f[A_n]$; then all $A_n$ are compact non-empty, and the $A_n$ are decreasing. Try to show that $A = \cap_n A_n$, which is also compact and non-empty, satisfies $f[A] = A$.

Another non-constructive way to show this is to consider the poset $\mathcal{P} = \{ A \subset X \mid A, \mbox{closed, non-empty and } f[A] \subset A \}$, ordered under reverse inclusion. Then an upper bound for a chain from $\mathcal{P}$ is the (non-empty) intersection, and a maximal element (by Zorn one exists) is a set $A$ with $f[A] = A$.

  • $\begingroup$ That's a good proof, but I don't need X (or H) Hausdorff. $\endgroup$ – cap Apr 25 '12 at 6:05
  • $\begingroup$ I assume you are in $\mathbb{R}$ and you already said your H is compact, do you know $\mathbb{R}$ is hausdorff space? $\endgroup$ – Marso Apr 25 '12 at 6:13
  • $\begingroup$ H is simply a metric space; no assumptions of R. $\endgroup$ – cap Apr 25 '12 at 6:15
  • 1
    $\begingroup$ do you know how to prove any metric space is hausdorff? $\endgroup$ – Marso Apr 25 '12 at 6:16
  • 3
    $\begingroup$ $A$ will be all of $X$ whenever $f$ is surjective. $\endgroup$ – Chris Eagle Apr 25 '12 at 6:28

In general, the answer is no. For example, let $H$ be a two-point space and $\omega$ be the function that swaps the two points.

  • 1
    $\begingroup$ Strictly speaking, such function fixes the whole $H$. $\endgroup$ – Vadim Nov 24 '14 at 22:41
  1. Here is an example showing that the assumption that $H$ is Hausdorff is needed.

Let $H$ be the set of natural numbers $1,2,3,\ldots$ in the cofinite topology (the topology in which a subset is closed iff it is finite or the whole space).

And consider $f(n)=n+1$.

Here are some facts:

a) The space is $T_1$ (in fact, the minimal $T_1$-space on this set) but not Hausdorff.

b) Every subset, including the whole space, is compact (since the space is not Hausdorff, a compact subset does not have to be closed).

c) $f$ is continuous (the preimage of every closed set, which is either the whole space or a finite subset, is either the whole space or a finite, hence closed, subset).

d) Obviously, there is no subset $A$ such that $f(A)=A$ (the only sets $A$ such that $f(A)\subseteq A$ are inductive sets $H_n=\{n,n+1,n+2,\ldots\}$, but $f(H_n)=H_{n+1}\subsetneq H_n$).

  1. However, as was mentioned by Une Femme Douce, if $H$ is compact Hausdorff, then for every continuous $f$ there is some $A$ such that $f(A)=A$. This is much easier to prove for a metric space, which seems to be the case after your comments.

Consider $A_0=H$, $A_n=f(A_{n-1})$. This is a monotonically decreasing sequence of compact (as images of compact sets under a continuous function), and hence, closed subsets. Their intersection $A=\cap_n A_n$ is non-empty.

Now, $f(A)=f(\cap_n A_n)\subseteq\cap_n f(A_n)=A$. Vice versa, suppose $z\in A$. Then for every $n$, there is $x_n\in A_n$ such that $f(x_n)=z\in A_{n+1}$. Since the space is compact and metric, it is sequentially compact, and there is a subsequence converging to some point $x$, and $x\in A$ (otherewise it is in some open set $H-A_j$ and no subsequence converges to it). Moreover, since $f$ is continuous and the space is Hausdorff, $f^{-1}(z)$ is closed and every $x_n\in f^{-1}(z)$, hence, $x\in f^{-1}(z)$, i.e. $f(x)=z$ and $z\in f(A)$. Overall, $A\subseteq f(A)$ and $A=f(A)$.

  1. In the general case, we don't even need the assumption that $f$ is continuous, all we need that $f$ maps compact subsets to compact subsets. But the proof is different.
  • $\begingroup$ Is it obvious to see that the space you constructed is compact? Unless I'm doing something wrong, it seems that the covering given by $1\cup \{2n\}_n, \{3n\}, \{5n\}, \{7n\}, \{11n\}$, etc. has no finite covering. $\endgroup$ – Tommy Tang Mar 26 '16 at 0:03
  • $\begingroup$ @TommyTang The sets you consider are not open in the finite complement topology. In the finite complement topology, for every set $A$, if an open set $U$ intersects it, then there are no more than a finite number of points left uncovered, so for each such point you can choose one open set from the covering, and you get a finite number of sets covering $A$. $\endgroup$ – Vadim Mar 26 '16 at 7:07
  • $\begingroup$ Oh oops. Thanks! $\endgroup$ – Tommy Tang Mar 26 '16 at 16:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.