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Why $\mathbb{Z}$ with $p$-adic topology is non-discrete?

Note1 : discrete : each singleton is an open set.

Note2 : Let the topology $\tau$ on $\mathbb{Z}_p$ be defined by taking as a basis all sets of the form $\{n + \lambda p^a \ ; \ \lambda \in \mathbb{Z}_p \& \ a \in \mathbb{N} \}$. Then $\mathbb{Z}_p$ is a compactification of $\mathbb{Z}$, under the derived topology (it is not a compactification of $\mathbb{Z}$ with its usual topology). The relative topology on $\mathbb{Z}$ as a subset of $\mathbb{Z}_p$ is called the $p$-adic topology on $\mathbb{Z}$.

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What is your definition of "discrete"? – Henning Makholm Oct 9 '12 at 19:31
And your definition of the $p$-adic topology? – Chris Eagle Oct 9 '12 at 19:43
what is $n$ in $\{n + \lambda p^a \ ; \ \lambda \in \mathbb{Z}_p \& \ a \in \mathbb{N} \}$? – Relative0 Nov 18 '14 at 13:08
up vote 3 down vote accepted

The $p$-adic topology on $\mathbb{Z}$ is the topology induced by the norm $|x|_p := p^{-v_p(x)}$, where $v_p$ is the $p$-adic valuation of $x$, that is, the multiplicity of $p$ in the factorization of $x$ as product of primes. Any open neighbourhood of $0$ contains a set of the form $U = \{x \in \mathbb{Z}\ |\ |x|_p < \varepsilon\}$ for some $\varepsilon > 0$. Clearly $x \in U$ if and only if $v_p(x) > - \log_p(\varepsilon)$, i.e. $x \in p^n \mathbb{Z}$ where $n$ is the cealing of $- \log_p(\varepsilon)$. In other words $p^n \mathbb{Z} \subseteq U$. In particular the open neighbourhoods of $0$ are infinite, so this topology is not discrete.

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In $p$-adic topology, two numbers are "close" if their difference is divisible by a high power of $p$ - the higher the better. So for example $3, 5, 9, 17, 33, 65, \ldots, 2^n+1,\ldots$ may look like diverging (which it is in the usual topology of $\mathbb Z$) but $2$-adically it converges to $1$ and the members of the sequence come arbitrarily close to $1$. This would not be possible in discrete topology.

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