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Statement

Let be $X$ a topological space that has a open-closed base $\mathcal{B}$. So if $X$ is such that $|X|>1$ then it is not locally connected.

Proof. If $X$ was locally connected then for any $x\in X$ there exist a connected neighborhood $V$ of $x$ and so there must exist $B\in\mathcal{B}$ such that $x\in B\subseteq V$ and so, since $B$ is open and closed, $B$ and $V\setminus B$ is two open disjoint set of $V$ such that $B\cup V\setminus B=V$, that means that $V$ is not connected.

Clearly the proof is correct iff $B\subset V$: indeed only in this case it follow that $V\setminus B\neq\varnothing$. Anyway I see that any discrete space $X$ is a space such that have a open-closed base and it is locally connected: so it seems that the statement is false; however my text says the contrary so I ask if we add other hypothesis then the statement is true. So how to prove the statement? Could someone help me, please?

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For each $x\in X$ let $\mathscr{B}(x)=\{B\in\mathscr{B}:x\in B\}$; it’s enough to add the requirement that there be a point $x_0\in X$ such that $\bigcap\mathscr{B}(x_0)$ is not open. (This is automatic if, for instance, $X$ is $T_1$ but not discrete.)

Let $H=\bigcap\mathscr{B}(x_0)$, and let $V$ be an open nbhd of $x_0$; then $H\subsetneqq V$, so fix $x\in V\setminus H$. There are $B_0,B_1\in\mathscr{B}(x_0)$ such that $B_0\subseteq V$ and $x\notin B_1$. Let $U=B_0\cap B_1$; then $U$ is clopen, and $x_0\in U\subseteq V\setminus\{x\}$, so $U$ and $V\setminus U$ are disjoint, non-empty open subsets of $V$ whose union is $V$.

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    $\begingroup$ @AntonioMariaDiMauro: I am not assuming that $X$ is $T_1$. $\mathscr{B}(x_0)$ is a local base at $x_0$, and $V$ is an open nbhd of $x_0$, so there must be some $B_0\in\mathscr{B}(x_0)$ such that $x_0\in B\subseteq V$. If $x\notin\bigcap\mathscr{B}(x_0)$, so there must be some $B_1\in\mathscr{B}(x_0)$ such that $x\notin B_1$. $B_0$ and $B_1$ are clopen, so their intersection is clopen. Finally, $V$ is an arbitrary open nbhd of $x$, and we’ve shown that it’s not connected, so $x$ has no connected open nbhd. $\endgroup$ Apr 16, 2020 at 20:36
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    $\begingroup$ @AntonioMariaDiMauro: Because they’re members of the clopen base $\mathscr{B}$: remember that $\mathscr{B}(x_0)$ is just the collection of members of $\mathscr{B}$ that contain $x_0$, so they’re all open. $\endgroup$ Apr 16, 2020 at 20:43
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    $\begingroup$ @AntonioMariaDiMauro: I’d forgotten that although weak local connectedness (your property) and local connectedness (my property) are not equivalent at a point, they are equivalent globally. You know from Prop. 8.23 that a space is weakly locally connected iff components of open sets are open. Now just prove that a space is locally connected iff components of open sets are open; it’s a slightly simpler version of the same argument. It follows that the two notions of local connectedness are equivalent, and you’re done. $\endgroup$ Apr 17, 2020 at 16:15
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    $\begingroup$ @AntonioMariaDiMauro: Imitate the proof of $8.23$. If $X$ is loc. conn., $U$ is open in $X$, and $C$ is a component of $U$, let $x\in C$; then $x$ has a conn. open nbhd $B_x$ such that $x\in B_x\subseteq U$, and since $B_x$ is conn., we must have $B_x\subseteq C$. Thus, $C=\bigcup_{x\in C}B_x$ is open. Conversely, suppose that components of open sets are open, let $U$ be open, and let $x\in U$; the component of $U$ containing $x$ is a conn. open nbhd of $x$ contained in $U$, so the conn. open nbhds of $x$ are a local base at $x$, and $X$ is loc. conn. $\endgroup$ Apr 17, 2020 at 17:01
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    $\begingroup$ @AntonioMariaDiMauro: Because that’s part of the definition of component. A component of a set $A$ is a maximal connected subset of $A$. Singletons are always connected, so the empty set is never a maximal connected subset of any non-empty set and hence never a component. (And we don’t generally talk about components of $\varnothing$.) $\endgroup$ Apr 17, 2020 at 17:36

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