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If $p$ and $q$ are distinct prime number, are $\mathbb{Q}_p$ and $\mathbb{Q}_q$ homeomorphic as topological space?

Frankenstein
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1 Answers1

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Yes. Let $X$ be the product of countably infinitely many copies of the discrete two-point spaces (i.e., a Cantor set), and let $x\in X$ be any point; then $\Bbb Q_p$ and $\Bbb Q_q$ are both homeomorphic to $X\setminus\{x\}$.

If you prefer, you can start with the middle-thirds Cantor set $C$ and delete any one point: they are homeomorphic to $C\setminus\{1\}$, for instance.

Brian M. Scott
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  • But there isn't any homeomorphism which keep fixed $\mathbb{Q}$, isn't it? (The sequence $(p^n)_{n \in \mathbb{N}}$ should be a counterexample) – Frankenstein Aug 20 '13 at 18:21
  • Is it possible to give some hints or references for the homomorphism from $\mathbb Q_p$ to $X$ minus one point ? Thanks in advance. – Cantlog Aug 25 '13 at 15:59
  • @Cantlog: Using the valuation metric one can show fairly easily that $\Bbb Z_p$ is homeomorphic to ${0,\dots,p-1}^{\Bbb N}$, where ${0,\dots,p-1}$ has the discrete topology; this is a Cantor set. (See my answer to this question.) It’s also $${x\in\Bbb Q_p:|x|p\le 1}={x\in\Bbb Q_p:|x|_p<p^{1/2}};,$$ which is clopen in $\Bbb Q_p$. For each $n\in\Bbb N$, $p^{-n}\Bbb Z_p$ is a clopen subset of $\Bbb Q_p$ homeomorphic to $\Bbb Z_p$, and $\Bbb Q_p=\bigcup{n\in\Bbb N}p^{-n}\Bbb Z_p$, so $\Bbb Q_p$ is homeomorphic to $\Bbb N\times\Bbb Z_p$, ... – Brian M. Scott Aug 25 '13 at 18:47
  • ... and hence to $\Bbb N\times C$, where $C$ is the middle-thirds Cantor set. Let $C_0=C\setminus{0}$. Then $C_0$ is the union of the sets $C\cap\left[\frac23,1\right]$, $C\cap\left[\frac29,\frac13\right]$, $C\cap\left[\frac2{27},\frac19\right]$, and so on, each of which is homeomorphic to $C$. Thus, $C_0$ is also homeomorphic to $\Bbb N\times C$ and hence to $\Bbb Q_p$. – Brian M. Scott Aug 25 '13 at 18:51
  • Great, thanks ! – Cantlog Aug 25 '13 at 22:11
  • @Cantlog: You’re welcome! – Brian M. Scott Aug 25 '13 at 22:30