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Recall that a topological space $X$ has the fixed point property (FPP) if any continuous function $f: X\to X$ has a fixed point.

Is the notion of FPP for topological spaces an absolute notion? More precisely:

Question. Is it consistent that a topological space $X$ has the FPP in $V$, but it does not have the FPP in some (cardinal preserving) generic extension of $V$?

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Does the (ground model) unit interval have FPP after adding a Cohen real? –  Joel David Hamkins Mar 16 '14 at 14:01
Perhaps one can make a counterexample space with a measurable cardinal after Prikry forcing, since this is cardinal preserving, but the new cofinality $\omega$ might open the door for a new continuous function without a fixed point. But I'm not sure what the space should be... –  Joel David Hamkins Mar 16 '14 at 14:44

2 Answers 2

up vote 15 down vote accepted

The answer is that the FPP is not absolute, and indeed, even the unit interval loses the FPP in a forcing extension. The unit interval famously has the FPP, but I claim that in any forcing extension having a new real, such as the forcing extension $V[c]$ obtained by adding a Cohen real, which preserves cardinals, the ground model unit interval no longer has the FPP.

To see this, let $X=I^V$ be the unit interval of $V$, considered as a topological space in $V[c]$. Let $a_n\to c$ be an increasing sequence of rational numbers converging to $c$ from below, and $b_n\to c$ from above. Let $f:X\to X$ be the piece-wise linear increasing function that takes the interval $[a_n,a_{n+1}]\to [a_{n+1},a_{n+2}]$ and similarly $[b_{n+1},b_n]\to [b_{n+2},b_{n+1}]$. This function takes ground-model reals to ground-model reals, but below $c$, it lies above the diagonal and above $c$, it lies below the diagonal. So it has no fixed point in $X$. (Meanwhile, it has a natural extension to the unit interval in $V[c]$, having fixed point $c$.)

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Thank you for your very nice answer. –  Mohammad Golshani Mar 17 '14 at 3:17

(Ought to be a comment, but too long:) One possible response to Joel's answer is to ask whether there is a definition of a topological space whose FPPness can be altered by forcing. That is, in the unit interval case, we do have $V[G]\models\neg$"[0, 1]$^V$ has FPP," but $V[G]$ does satisfy "[0, 1] has FPP." So it's reasonable to ask whether there is definition of a topological space which is a counterexample.

Arnold Miller (http://arxiv.org/abs/0806.1957) has shown that there is (relative to $Con(ZFC)$) a model $V$ of $ZF$ in which there is a Borel (in fact, $F_{\sigma\delta}$) strictly Dedekind-finite set $D$ of reals. (Here "strictly Dedekind-finite" means "infinite, but without a proper injective self-map.") A Borel code $\alpha$ for $D$ gives a reasonable "definition" of $D$, and this definition yields an infinite set of reals in any generic extension of $V$; now let $\mathcal{T}$ be the definition, "the discrete topology on the Borel set coded by $\alpha$". Clearly once we add a well-ordering of $\mathbb{R}$ the topology defined by $\mathcal{T}$ will not have FPP, but it isn't obvious to me that $\mathcal{T}$ has FPP in $V$.

Anyways, something along these lines might be interesting.

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One could consider "the unit interval of $L$" as a definition, and my example would still be an example of the kind you seek, provided that $\mathbb{R}^V\subset L$. –  Joel David Hamkins Mar 17 '14 at 1:29
That's a good point - although that definition is (?) $\Delta^1_2$, whereas if something like Miller's Borel strictly Dedekind finite set lacked FPP, then we'd have a definition of much lower complexity, and probably(?) close to minimal possible complexity - at least, among topological spaces whose underlying set is a set of reals. –  Noah Schweber Mar 17 '14 at 1:35
(Although I'll admit to an ulterior motive: any opportunity to mention Miller's bizarre Borel sets is a good one, to my mind.) –  Noah Schweber Mar 17 '14 at 1:37

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