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I have seen two a priori different definitions of locally connected space :

1) For all point $x$, and neighborhood $V$ of $x$, there is a connected neighborhood $C$ of $x$ such that $C\subseteq V$

2) For all point $x$, and open set $U$ containing $x$, there is an open connected neighborhood $O$ of $x$ such that $O\subseteq U$.

I imagine those two definitions are identical, but I don't see why.

The same thing for locally path-connected spaces, we can either use open neighborhoods or just neighborhoods, are the two definitions the same?

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  • $\begingroup$ Write down the definition of neighborhood. Clearly one has $(1) \implies (2)$ as if $U$ is an open set containing $x$, then $U$ is a neighborhood of $x$. $(2) \implies (1)$ comes from the definition. $\endgroup$ – Hermès Oct 1 '16 at 16:56
  • $\begingroup$ I dont understand the $1 \to 2$ part. If I consider $U$ as my neighborhood, I only get a connected neighborhood $C$ included in $U$, and nothing tells me that $C$ is open, which is what I want for 2. $\endgroup$ – Jon-S Oct 1 '16 at 17:03
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By definition a neighborhood of a point $x$ is a set $I$ that contains an open set $A$ such that $x\in A$. So 1) $\Rightarrow$ 2) by considering $U$ as your neighborhood; 2) $\Rightarrow$ 1) by considering the open set of the definition of neighborhood.

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    $\begingroup$ I dont understand the $1 \to 2$ part. If I consider $U$ as my neighborhood, I only get a connected neighborhood $C$ included in $U$, and nothing tells me that $C$ is open, which is what I want for 2. $\endgroup$ – Jon-S Oct 1 '16 at 17:02
  • $\begingroup$ $C$ contains by definition an open set $L$ that contains $x$; it is a neighborhood by definition, it is open and it is connected because, if it was disjoint union of open sets $A,B$ you can consider the one that contains $x$. $\endgroup$ – Giuseppe Bargagnati Oct 1 '16 at 17:24
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    $\begingroup$ @Jon-S: Yes, that’s the correct approach. There’s a proof here. $\endgroup$ – Brian M. Scott Oct 1 '16 at 23:06
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    $\begingroup$ @Brian M. Scott : Indeed that answers the question for connectedness, thanks. Is there the same type of result for path-connectedness ? $\endgroup$ – Jon-S Oct 1 '16 at 23:47
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    $\begingroup$ @Jon-S: Essentially the same argument should work. $\endgroup$ – Brian M. Scott Oct 1 '16 at 23:52

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