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Let $p$ be a prime number. Let $n,m \geq 1$ be such that the topological spaces $\mathbb{Q}_p^n$ and $\mathbb{Q}_p^m$ are homeomorphic. Can we conclude $n=m$?

For $\mathbb{Z}_p$ it's false: In fact, Brouwer's theorem implies that $\mathbb{Z}_p$ is homeomorphic to the Cantor set $C$, which of course satisfies $C^n \cong C^m$ for all $n,m$.

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$\mathbb{Q}_p$ is homeomorphic to a countable direct sum of copies of the Cantor set $C$. Indeed, because the valuation is discrete, for each $n \geq 1$ the "annulus"

$A_n =$ {$x \in \mathbb{Q}_p \ | \ p^{n-1} < ||x|| \leq p^{n}$}

is closed and homeomorphic to the Cantor set $C$. (Take of course $A_0 = \mathbb{Z}_p$.)

Since as you observed above, $C \times C \cong C$, it follows that $\mathbb{Q}_p^n$ and $\mathbb{Q}_p^m$ are homeomorphic for all $m, \ n \in \mathbb{Z}^+$ (and the homeomorphism type is independent of $p$).

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  • $\begingroup$ A topologist would probably say "disjoint union" rather than "direct sum". $\endgroup$
    – Gerald Edgar
    Commented Dec 31, 2009 at 15:19
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    $\begingroup$ That's true. So probably I'm not a topologist, then. (Seriously, the poster has shown great familiarity with categorical language and ideas, so I'm sure he understands the "direct sum" and would conjecture that he thinks of it that way himself.) $\endgroup$
    – Pete L. Clark
    Commented Jan 1, 2010 at 10:28

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