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

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3 Answers 3

up vote 2 down vote accepted

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 $f[A] = A$.

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

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Strictly speaking, such function fixes the whole $H$. –  Vadim yesterday
  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.
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