# If $X$ has a open-closed base $\mathcal{B}$ and if $|X|>1$ then $X$ is not locally connected

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?

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 $$V$$.
• @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. Apr 16, 2020 at 20:36
• @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. Apr 16, 2020 at 20:43
• @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. Apr 17, 2020 at 17:01
• @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 $\varnothing$.) Apr 17, 2020 at 17:36