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We know that all connected subsets of $\mathbb{R}$( with the usual topology) has no empty interior. I would like to know if this fact remains true for a general connected topological space with the Lebesgue covering dimension equal 1.

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If $C \subset [0,1]$ is the Cantor set, how about $([0,1] \times C) \cup (\{0\} \times [0,1])$ as a subset of $\mathbb{R}^2$? It's connected and one-dimensional, but $[0,1] \times \{0\}$ is nowhere dense.

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  • $\begingroup$ @ NateEldredge The real line has a basis formed by connected open sets, while your set does not have a such basis. Maybe with this additional assumption (that is, the topological space has a basis formed by connected sets), the statement is true? $\endgroup$
    – Didi
    Commented Nov 16, 2016 at 21:11

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