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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!

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What topologies are given for the irrationals and the rationals? The subspace topology from their inclusions in the real line? – Shaun Ault Sep 2 '12 at 21:38
@Shaun If someone doesn't say what topology a subset of $\Bbb R$ has, you should assume it's the usual one. – MJD Sep 2 '12 at 21:49
@ShaunAult : Those are linearly ordered sets and there's an order topology. I'm pretty sure that's the same as the subspace topology that you mention, but it seems simpler to put it that way. – Michael Hardy Sep 2 '12 at 22:13
up vote 6 down vote accepted

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.

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A slightly different way of putting it: If $\mathbb{P}$ were homeomorphic to $\mathbb{P} \times \mathbb{Q} = \bigcup_{q \in \mathbb{Q}} \mathbb{P} \times \{q\}$ it would be a countable union of nowhere dense sets, contradicting Baire. – t.b. Sep 2 '12 at 21:52
@t.b.'s remark provides a proof that doesn't require showing that $\mathbb P$ is topologically complete: all you have to note is that if $\mathbb P$ was a countable union of nowhere dense closed (in $\mathbb P$) sets $S_j$, ${\mathbb R}$ would be as well. Just take the closures (in $\mathbb R$) of $S_j$ and the singletons $\{q\}$ for $q \in \mathbb Q$. – Robert Israel Sep 2 '12 at 22:01
Is it clear that $\Bbb P$ is topologically complete? I didn't know that, and I can't think of a proof offhand. – MJD Sep 2 '12 at 22:01
@t.b. That should go on an answer rather than a comment! – Asaf Karagila Sep 2 '12 at 22:06
@MJD: It’s well enough known that it’s part of my mental furniture. It follows from the fact that a subset of a complete metric space $X$ is topologically complete iff it’s a $G_\delta$ in $X$, which is standard fare. – Brian M. Scott Sep 2 '12 at 22:09

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