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Let $(X,\mathcal{T})$ be a connected topological space. Let $\mathcal{T}'$ be the topology on $X$ that is generated by the collection of connected open sets in $(X,\mathcal{T})$. That is, the connected open sets in $(X,\mathcal{T})$ form a subbasis for $\mathcal{T}'$. Is it necessarily true that $(X,\mathcal{T})$ and $(X,\mathcal{T}')$ have the same collection of connected open sets?

It is true if $(X,\mathcal{T})$ is locally connected. To me it feels like the statement should be false for some space that is not locally connected, but the usual examples of spaces that are not locally connected are not counterexamples. This question is partly motivated by Connected topological space $X$ such that $\emptyset, X$ are the only open connected subsets , which also doesn't seem to provide a counterexample for my statement.

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    $\begingroup$ You probably want some additional assumptions. As is, you can just take $X$ to be the Cantor set, then there’s no connected open set apart from $\varnothing$, so $\mathcal{T}’$ would be indiscrete, in which case you have two connected open sets, namely $\varnothing$ and $X$. $\endgroup$
    – David Gao
    Commented Dec 3 at 9:31
  • $\begingroup$ @DavidGao You're right. At least the connectivity of $(X,\mathcal{T})$ would be needed. I missed that. Thanks! $\endgroup$ Commented Dec 3 at 10:23

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Take $\mathbb{Q}$ and add two points at infinity, $\infty_1$ and $\infty_2$, topologizing so that $\mathbb{Q} \cup \{\infty_i\}$ is the only open proper subset containing $\infty_i$ for $i \in \{1,2\}$. The space is now connected, but in the coarse topology $\mathbb{Q}$ is connected and open.

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  • $\begingroup$ This shows that we should probably add a Hausdorff requirement $\endgroup$
    – Smiley1000
    Commented Dec 3 at 16:01

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