0
$\begingroup$

This may be a stupid question.But I am stuck with it.Is Q_p(the p-adic) connected under the usual topology?I was confounded with this problem while trying to construct a counter-example related to my master's thesis.

$\endgroup$

closed as off topic by George Lowther, Chandan Singh Dalawat, user9198, Tony Huynh, Marc Palm Dec 19 '12 at 10:03

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Both Wikipedia and Google know the answer to this (very elementary) question. If you don't find it there or by yourself, you should try math.stackexchange.com $\endgroup$ – Olivier Dec 19 '12 at 9:34
  • 2
    $\begingroup$ No, $\mathbf{Q}_p$ is about as far from being connected as possible: it's totally disconnected - the maximal connected subsets are points. $\endgroup$ – David Loeffler Dec 19 '12 at 9:35
  • 2
    $\begingroup$ $\mathbb{Q}_p$ is a punctured Cantor set, meaning that it is homeomorphic to a Cantor set minus one point. $\endgroup$ – Lee Mosher Dec 19 '12 at 14:39
1
$\begingroup$

Any non-empty open can be written as a disjoint union of opens ; for example $\mathbb{Z}_p=\cup(a+p\mathbb{Z}_p)$ where $a$ runs through $\{0...(p-1)\}$. Those spaces are said totally disconnected.

$\endgroup$
  • 1
    $\begingroup$ actually this happens for any topological group admitting a base of neighborhoods given by subgroups $\endgroup$ – Simone Virili Dec 19 '12 at 11:33

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