Geometric realization of the product of two simplicial sets Let $X$ and $Y$ be two simplicial sets.  Then there is a natural continuous map $\lvert X \times Y \rvert \to \lvert X \rvert \times \lvert Y \rvert$, where the right-hand side is given the product topology.  I've often seen the assertion in the literature that this map is not necessarily a homeomorphism, hence the need to work in some more convenient category.  
What is an example for which this map fails to be a homeomorphism?  I suspect that the simplicial sets involved in such a counterexample will be quite "large", but I haven't seen enough cases of strange geometric realizations to come up with the counterexample on my own.  
 A: The following counterexample is apparently due to Dowker and is taken from the appendix of Hatcher's Algebraic Topology (p. 524-5).  Let $X=\bigvee_{f\in S} I_f$ be a wedge sum of continuum many $1$-simplices $I_f$, indexed by the set $S$ of all functions $\mathbb{N}\to(0,1]$.  Let $Y=\bigvee_{n\in\mathbb{N}} I_n$ be a wedge sum of countably infinitely many $1$-simplices $I_n$, indexed by the natural numbers.  Given $f\in S$ and $n\in\mathbb{N}$, define $p_{fn}=(f(n),f(n))\in |I_f\times I_n|\subset|X\times Y|$, where we identify $|I_f\times I_n|$ with $[0,1]^2$, where $0$ corresponds to the wedge point in both coordinates.  Let $P\subset|X\times Y|$ denote the set of these points $p_{fn}=(f(n),f(n))$ for all $f$ and $n$.  Then $P$ is closed in $|X\times Y|$ (its intersection with any simplex consists of at most one point).
However, $P$ is not closed as a subset of $|X|\times |Y|$ with the product topology.  Indeed, letting $0$ denote the wedge point of both $|X|$ and $|Y|$, I claim that $(0,0)$ is in the closure of $P$ in $|X|\times |Y|$.  To prove this, let $U$ be a neighborhood of $0$ in $|X|$ and $V$ be a neighborhood of $0$ in $|Y|$.  For each $n\in\mathbb{N}$, there is some $t_n\in(0,1]$ such that $V\cap |I_n|$ contains $[0,t_n)$, identifying $|I_n|$ with $[0,1]$.  We may assume the sequence $(t_n)$ converges to $0$ (if it doesn't, just make the numbers $t_n$ smaller).  Now define $f:\mathbb{N}\to(0,1]$ by $f(n)=t_n/2$.  Then $U\cap |I_f|$ contains $[0,s)$ for some $s>0$, identifying $|I_f|$ with $[0,1]$.  Now choose $n$ sufficiently large so that $f(n)<s$.  We then find that the point $p_{fn}=(f(n),f(n))\in |I_f\times I_n|$ is contained in $U\times V$.  That is, $U\times V$ intersects $P$.  Since $U$ and $V$ were arbitrary neighborhoods of $0$, this means $(0,0)$ is in the closure of $P$ with respect to the product topology.
