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Let $X$ be a topological space; then we say that $X$ is locally connected at $x$ if $x$ admits a neighborhood basis of open, connected sets. In this sense, a space is locally connected iff it is locally connected at every point.

Consider the following.

  • It is easy to see that a space that is both locally connected and totally disconnected must be discrete.
  • Cantor's leaking tent is an example of a space which is totally disconnected but not locally connected at any point.
  • Several spaces are locally connected but not totally disconnected.

I've thought about spaces not locally connected at any point, since they are necessarily "much weirder" than the usual non-locally connected space. Perhaps this condition is equivalent to another one?

I suspect there must be a space which is neither locally connected at any point nor totally disconnected, but I haven't been able to produce or find an example.

Thank you in advance.

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How about this space? $$\{(x,y)\in\Bbb R^2\mid x\in\{\frac1n+\frac1m\mid n,m\in\Bbb N\}, y\in[0,1] \}$$

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  • $\begingroup$ By the way, are you sure this works? I might be missing something, but it seems to me that this space is locally connected at the points of the form $(2,y)$? $\endgroup$ – Dejan Govc May 26 '15 at 10:29
  • $\begingroup$ @DejanGovc I think you're right. I guess there's no way to fix it either. Hmph $\endgroup$ – Gregory Grant May 26 '15 at 10:34
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Here's another example: $$X=(\mathbb R\times\{0\})\cup(\mathbb Q\times\mathbb Q).$$ Any neighborhood of any point is disconnected, but $\mathbb R\times\{0\}$ is a connected component, so the space is not totally disconnected.

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    $\begingroup$ In that spirit, how about just $\mathbb R\times\mathbb Q$? $\endgroup$ – Gregory Grant May 26 '15 at 10:16
  • $\begingroup$ @GregoryGrant: that's even better. $\endgroup$ – Dejan Govc May 26 '15 at 10:17
  • $\begingroup$ Yeah, $\mathbb R\times\mathbb Q$ is a nice clean answer to this question. $\endgroup$ – Gregory Grant May 26 '15 at 10:18

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