Suppose $X$ is a topological space that is locally conneted and let $O$ be an open subset of $X$. Then we want to show that $O$ is also locally connected.

Let $p\in O$ chosen arbitrarily, then there exists a open set $U\subset O$ and a connected open set $V\subset X$, both of which containing $p$. Now their intersection $W=U\cap V$ is open and contains $p$, therefore $W$ is a neighborhood of $p$ and $W\subset U\subset O$. If $W$ is connected then we are done. Now, suppose that $W$ is not connected hence can be written as a disjoint union of two clopen sets $A$ and $B$. Then $A\cup B=W = U\cap V\subset V$. But since $V$ is connected, then the only clopen subsets of $V$ are $\phi$ and $V$ itself. We get the following cases:

  • Case 1: $A=B=\phi$, we have a contradiction since $p\in W=A\cup B$.
  • Case 2: $A=V$ or $B=V$, then $V= A\cup B=W=U\cap V\subset U\subset O$ and and since $p\in V\subset U \subset O$, therefore $O$ is locally connected since the choice of $p$ was arbitrary.

Is the proof okay? Did I make a mistake somewhere?

  • $\begingroup$ connected or compact? you better edit the question. $\endgroup$ – sha Aug 18 '15 at 21:26
  • $\begingroup$ replace compact with connected :) $\endgroup$ – Adel Saleh Aug 18 '15 at 21:32

in the definition of locally connected as i know, it is to say that for every $x\in X$, every open set $U$ such that $x\in U$ has an open and connected set $W$ such that $x\in W\subseteq U$

so i believe that if you will exchange in your proof that

there exists $U\subseteq \cal O$


for every $U\subseteq \cal O$ open in $\cal O$

you will get it.

  • $\begingroup$ You sure it still works for all $U$? $\endgroup$ – Adel Saleh Aug 18 '15 at 21:52
  • $\begingroup$ for all $U$ such that $p\in U$. $\endgroup$ – sha Aug 18 '15 at 22:01
  • $\begingroup$ say, let $U$ be an open set in $\cal O$ such that $p\in U$ and the rest will hold for that $U$ and since $U$ is arbitrary then it holds for every such $U$ $\endgroup$ – sha Aug 18 '15 at 22:02
  • $\begingroup$ Thanks a lot! The proof works! I checked the definition again. $\endgroup$ – Adel Saleh Aug 18 '15 at 22:08
  • $\begingroup$ glad i could help! you did had most of the proof $\endgroup$ – sha Aug 18 '15 at 22:10

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