This may be a stupid question.But I am stuck with it.Is Q_p(the padic) connected under the usual topology?I was confounded with this problem while trying to construct a counterexample related to my master's thesis.
closed as off topic by George Lowther, Chandan Singh Dalawat, user9198, Tony Huynh, Marc Palm Dec 19 '12 at 10:03
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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
Any nonempty 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 $\{0...(p1)\}$. Those spaces are said totally disconnected.

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