Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb{Z}$.

share|improve this question
4  
What is your definition of "discrete"? –  Henning Makholm Oct 9 '12 at 19:31
1  
And your definition of the $p$-adic topology? –  Chris Eagle Oct 9 '12 at 19:43
add comment

2 Answers

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.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.