TITLE: Would this space be homeomorphic to the set of irrationals?
QUESTION [5 upvotes]: I've been reviewing various problems dealing with interesting homeomorphisms, and I came across this one.

Is the product of the space of irrationals and the space of rationals homeomorphic to the space of irrationals?

I haven't been able to make any progress on this one. Can anyone help? Thank you!

REPLY [7 votes]: Let $\Bbb P$ be the space of irrationals; it is topologically complete, meaning that it has a compatible complete metric. Suppose that $\Bbb P\times\Bbb Q$ were homeomorphic to $\Bbb P$; then it would be topologically complete, so each of its closed subspaces would also be topologically complete. But $\Bbb P\times\Bbb Q$ certainly has closed subspaces homeomorphic to $\Bbb Q$, which is not topologically complete, so $\Bbb P\times\Bbb Q\not\cong\Bbb P$.
Added: In this paper Jan van Mill proved that $\Bbb P\times\Bbb Q$ is the unique space (up to homeomorphism) that can be written as an increasing union $\bigcup_{n\in\Bbb N}F_n$ of closed sets such that for each $n\in\Bbb N$, $F_n$ is a copy of $\Bbb P$ that is nowhere dense in $F_{n+1}$. (To write $\Bbb P\times\Bbb Q$ this way, enumerate $\Bbb Q=\{q_n:n\in\Bbb N\}$, and let $F_n=\Bbb P\times\{q_k:k\le n\}$.) It’s a nice little exercise in the Baire category theorem to show that $\Bbb P$ cannot be written in this way.