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Let $X$ be a zero-dimensional subset of the plane $\mathbb R ^2$. Is $\mathbb R ^2\setminus X$ necessarily path-connected? I feel the answer must be yes but I need a reference. If it helps, assume $X$ is nowhere dense.

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If the zero-dimensional set $X$ is not closed, then the answer is "no".

To construct a suitable example, take any open bounded neighborhood $U\subset\mathbb R^2$ of zero, whose boundary $\partial U$ does not contain a topological copy of $[0,1]$. For example, for $U$ we can take a bounded connected component of the complement of the union of two suitable pseudoarcs in the plane. Then the set $$X=\mathbb R^2\setminus\{\vec a+\tfrac1n\partial U:\vec a\in\mathbb Q^2,\;n\in\mathbb N\}$$ will have the desired property: it is zero-dimensional and its complement $\mathbb R^2\setminus X$ does not contain a copy of $[0,1]$ (by the Baire Theorem) and hence is not path-connected.

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At least if $X$ is compact, the answer is yes. Indeed, by Corollary 2 of Theorem IV 3 in:

W. Hurewicz, H. Wallman, Dimension Theory. Princeton Mathematical Series, v. 4. Princeton University Press, Princeton, N. J., 1941,

compact sets separating points in $\mathbb{R}^{n+1}$ must have topological dimension $n$.

In particular, compact sets separating points in $\mathbb{R}^2$ must have (topological) dimension $1$ or $2$ so sets of dimension $0$ cannot separate points.

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  • $\begingroup$ You know, reading your answer makes me wonder which specific kind of topological dimension OP had in mind (but then again R^2 is polish so it doesn't really matter...) $\endgroup$
    – nomen
    Feb 3 '20 at 20:06

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