We know that all connected (not a singleton) subsets of $\mathbb{R}$ (with the usual topology) has no empty interior. This fact does not remains true for a general connected topological space with the Lebesgue covering dimension equal 1 as showed in Topological spaces with Lebesgue covering dimension 1. However, the topological space presented in that post does not have a basis of connected sets. My question is: If $X$ is a connected topological space with Lebesgue covering dimension 1 having a basis of connected subsets then all connected subset (not a singleton) has no empty interior?
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5
Unless I'm missing something, the answer is no: consider the space $\mathbb{R}'=\mathbb{R}\cup\{\alpha\}$, with topology generated by $$\{(a, b): a<b, 0\not\in (a, b)\}\cup\{(a, b)\cup\{\alpha\}: a<0<b\};$$ so we've duplicated $0$ to break $T_0$-ness. Then $\mathbb{R}'$ still has Lebesgue dimension $1$ and a basis of connected sets, but $\{0, \alpha\}$ is a connected non-singleton subset of $\mathbb{R}'$ with empty interior.
Maybe you want to restrict to $T_0$-spaces (or better)?
answered Nov 17, 2016 at 22:35
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$\begingroup$ @georglehner no, it's indiscrete - every open containing 0 contains a and vice versa. $\endgroup$ Commented Nov 17, 2016 at 23:13
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$\begingroup$ I misread thinking you meant the line with doubled origin. My bad. $\endgroup$ Commented Nov 17, 2016 at 23:16
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$\begingroup$ @NoahSchweber Do you think that it is true for metric space? I would like it were true for compact metric spaces. $\endgroup$– DidiCommented Nov 17, 2016 at 23:53
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5$\begingroup$ @Didi Attach a $[0,\tfrac1q]$ interval to each rational point $\tfrac pq$ in $[0,1]$. You get a compact metric space $K$. The set $[0,1]\subset K$ is connected, and has empty interior. $\endgroup$ Commented Nov 18, 2016 at 0:03
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4$\begingroup$ @AntonPetrunin I feel like that should really be a separate (and accepted!) answer - it's much better than mine. $\endgroup$ Commented Nov 18, 2016 at 0:13