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Let $X$ be a set and $\{\mathcal{T}_i \subset \mathcal{P}(X) : i \in I\}$ be an arbitrary set of locally connected topologies in $X$; it may or may not contain all of them and $I$ may be finite or infinite.

Is $\mathcal{S} = \bigcap \{\mathcal{T}_i : i \in I\}$ a locally connected topology in $X$?

Notes:

  • In the question Is intersection of refining locally connected topologies locally connected? it was (amazingly!) shown that the claim is true when the above set contains all locally connected topologies which refine a given topology.
  • $\mathcal{S}$ is a topology in $X$ (proof here)
  • If $(X, \mathcal{T})$ is finite or finitely generated, then $(X, \mathcal{T})$ is locally connected.
  • So a possible counter-example must have infinite $X$.
kaba
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1 Answers1

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Yes, this follows almost immediately from the answer to your previous question. Namely, let $\mathcal{S}'$ be the smallest locally connected topology that contains $\mathcal{S}$ (which exists by the answer to your previous question). Then each $\mathcal{T}_i$ contains $\mathcal{S}'$, since it is locally connected and contains $\mathcal{S}$. Thus $\mathcal{S}'\subseteq\mathcal{S}$ and so $\mathcal{S}'=\mathcal{S}$, so $\mathcal{S}$ is locally connected.

You can also prove this more directly as follows. Suppose $U$ is $\mathcal{S}$-open and let $C$ be an $\mathcal{S}$-connected component of $U$. Then for each $i$, $C$ is a union of $\mathcal{T}_i$-connected components of $U$, since $\mathcal{T}_i$ contains $\mathcal{S}$. Since $U$ is $\mathcal{T}_i$-open and $\mathcal{T}_i$ is locally connected, this implies $C$ is $\mathcal{T}_i$-open. Since $i$ was arbitrary, this means $C$ is $\mathcal{S}$-open.

Eric Wofsey
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  • Right you are. Thanks! The latter direct proof then of course works for the previous refinement problem also, and is simpler. Many times it seems that the component definition is more useful for local connectedness than the neighborhood definition. – kaba Aug 10 '22 at 12:51