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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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    @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.2012-09-02

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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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    @t.b.: I thought you might. I remember sitting down and working out something very similar when I first learned that the irrationals are homeomorphic to $\omega^\omega$.2012-09-02