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Why $\mathbb{Z}$ (group of integer numbers) with $p$-adic topology is a countable precompact metric group with a linear topology?

Note : Call a topological group $G$ linear (and its topology a linear group topology) if $G$ has a base of $e$ formed by open subgroups of $G$.

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Which of the 5 properties do you not believe? Only prcompact? –  Hagen von Eitzen Oct 5 '12 at 18:34
    
yes,just precompact. –  Wreza Shafaghi Oct 5 '12 at 20:18
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1 Answer 1

Given $\varepsilon>0$ find $k$ with $\frac1{p^k}<\varepsilon$. Then The open $\varepsilon$-balls around $0, 1, \ldots, p^k-1$ cover $\mathbb Z$. Hence precompactness.

Let $U$ be an open neighbourhood of $0$. Then some $\varepsilon$-ball around $0$ is contained in $U$. As above, this ball contains $p^k\mathbb Z$ for $k$ big enough, which is a subgroup. Hence linearity.

(Countable metric group should of course be clear)

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please more explain your reason about precompactness –  Wreza Shafaghi Oct 8 '12 at 14:12
    
I think the diffinition of precompact is : A subset in a topological space is precompact if its closure is compact –  Wreza Shafaghi Oct 8 '12 at 14:28
    
Or : A uniform space is pre-compact if and only if every net in has a Cauchy subnet. –  Wreza Shafaghi Oct 8 '12 at 14:29
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