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Give an example of a subset of $\mathbb R^2$ which is path connected, but no point has a local base of connected sets.

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  • $\begingroup$ Do you mean that no point has a local base of connected neighborhoods? $\endgroup$ – Stefan Hamcke Sep 20 '13 at 16:37
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$$I\times\{0\} \cup \left(\{0\}\cup\left\{\frac1n\middle|\ n\in\Bbb N\right\}\right)\times I$$

If you want a space where no point has a local base of connected sets, consider the following

Let $U_{n,q}$ denote the line segment joining $(n+q,n)$ to $(n+q,n+1)$, and let $R_{n,q}$ denote the segment between $(n+1,n+q)$ and $(n+2,n+q)$. Define $Z$ as the union $$Z=\bigcup \{R_{n,q}, U_{n,q}\mid n\in\Bbb Z,q\in\Bbb Q\cap I\}$$ This space is path-connected, but not locally connected at any point.

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  • $\begingroup$ Does anyone know how to enlarge the delimiter " $|$ " in the set notation? $\endgroup$ – Stefan Hamcke Sep 20 '13 at 16:03
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    $\begingroup$ \middle|. It's worth noting that this is the so-called comb space. $\endgroup$ – Ayman Hourieh Sep 20 '13 at 16:11
  • $\begingroup$ You're welcome! $\endgroup$ – Ayman Hourieh Sep 20 '13 at 16:13
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    $\begingroup$ @Mathemagician1234 My example is not homeomorphic to the comb space (for example comb space is compact). You can also take a cone on the rationals, or similar things. $\endgroup$ – ronno Sep 20 '13 at 16:19
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    $\begingroup$ excuse the statement was wrong! $\endgroup$ – Jarbas Dantas Silva Sep 20 '13 at 16:27
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$$\{(x,y) \in I \times I| x \in \Bbb{Q} \text{ or } y = 0\}$$

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    $\begingroup$ excuse the statement was wrong! $\endgroup$ – Jarbas Dantas Silva Sep 20 '13 at 16:28
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$[\displaystyle\cup_{(\ \in [0,1] \cap \mathbb{Q})\times 0)} (a(t-1)+p_1t; \ t \in (0,1), \ p_1=(0,1) \in \mathbb{R^2})] \bigcup [\displaystyle\cup_{(b \in 0 \times [0,1] \cap \mathbb{Q})} (b(t-1)+p_2t; \ t \in (0,1), \ p_2=(0,0) \in \mathbb{R^2})] $

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