I wanted to know if a space $(X, \tau)$ with $\tau$ the finite-closed topology, have the Fixed Point Property, that is if every continuous mapping $f$ of $(X, \tau)$ into itself has a fixed point.
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$\begingroup$ If $X$ is finite, then $(X,\tau)$ is discrete, and therefore it doesn't have the fixed point property unless $X=\{x\}$. $\endgroup$– user239203Nov 23, 2020 at 22:21
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1$\begingroup$ In that topology every bijection of $X$ to itself is continuous, and there are always bijections without fixed points unless $|X|=1$. $\endgroup$– Brian M. ScottNov 23, 2020 at 22:23
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1$\begingroup$ What is the "finite-closed topology"? The topology in which all proper closed sets are finite? If so, as @BrianM.Scott says, this is not true. (In fact, I'm pretty sure this is true for any topology which you can specify as "the" topology, without specifying the underlying set or any other data...) $\endgroup$– tomaszNov 23, 2020 at 22:31
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2$\begingroup$ @tomasz: I’m assuming that it’s an alternative name for the cofinite topology; I’m pretty sure that I’ve seen the term used that way. $\endgroup$– Brian M. ScottNov 23, 2020 at 22:41
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1$\begingroup$ @tomasz Derp, not my finest moment. $\endgroup$– Noah SchweberNov 23, 2020 at 23:13
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2 Answers
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The topology you have in mind - more commonly called the cofinite topology - looks a lot like the discrete topology in a few ways. One of these ways is the following:
Every self-bijection is continuous with respect to the cofinite topology.
The proof is basically immediate: the preimage of a cofinite set under an injective map must be cofinite.
Now if $X$ is any set with more than one element, there is a permutation of $X$ with no fixed points (exercise$^1$), so in fact no space with the cofinite topology which has more than one point has the fixed point property.
$^1$Amusingly, this actually requires a small amount of the axiom of choice: it is consistent with $\mathsf{ZF}$ (= set theory without choice) that there is a set with more than one element such that every self-bijection of the set has a fixed point. In fact it is consistent with $\mathsf{ZF}$ that there is an infinite set $A$ such that every self-bijection of $A$ fixes all but finitely many points! This is very difficult to prove however (as all $\mathsf{ZF}$-consistency results are), but Cohen's original model of $\mathsf{ZF+\neg AC}$ provides an example. In such a model we do indeed have infinite sets which, when equipped with the cofinite topology, have the fixed point property. But these are extremely pathological models.
answered Nov 23, 2020 at 22:39
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The natural numbers $\mathbb{N}$ with the cofinite( so that means finite sets are closed) topology does not have the fixed point property, since the $\sigma(n)=n+1$ map does not have fixed points, however takes open sets to open sets.
