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Suppose $A$ and $B$ are spaces of $X$ and $A$ and $B$　are locally connected and $A\cap B\neq \phi$,Must $A\cup B$ locally connected?

Part of my thought:Suppose $x\in A$, $ x\notin B$,for any open neighborhood of $x$ in $A\cup B$,$U$,there exists an open set in $X$ ,$V$,s.t. $U=(V\cap A)\cup(V\cap B)$.Considering open set in $A$,$V\cap A$,there exists a connected open connected neighborhood in $A$ contained in $V\cap A$.But this open connected neighborhood in $A$ may be not an open neighborhood in $A\cup B$. Can we disprove this statement from this detail?

Jack
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Let $X=\mathbb R^2$ and let $$A=\{(x,y):x\gt0\text{ and }y=\sin\frac1x\},$$ $$B=\{(x,y):x=0\text{ or }x=1\text{ or }y=2\}.$$ Then $A$ and $B$ are connected (which you did not require but perhaps intended) and locally connected, and $A\cap B=\{(1,\sin1)\}\ne\emptyset,$ but $A\cup B$ is not locally connected.

bof
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I think the following modification of the original statement is true: Let $\mathcal{A}$ be a collection of open locally connected sets of $X$ such that $\bigcup \mathcal{A} = X$. Then $X$ is locally connected.

Let $x \in X$ and $U$ be a neighborhood of $x$ in $X$. $x \in A$ for some $A$. $A \cap U \subset U$ is a neighborhood of $x$ in $A$, so there an open connected neighborhood $V \subset A \cap U$. $V$ is open and connected in $X$.

vincent wang
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