# Is the complement of a zero-dimensional subset of the plane path-connected?

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.

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.

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.

• 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...) Feb 3 '20 at 20:06