TITLE: Is the notion of fixed point property for topological spaces an absolute notion?
QUESTION [18 upvotes]: 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 $V$?

REPLY [6 votes]: (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.