Paths in co-Countable Subsets of the Plane

A rather simple question we can ask about any topological space is: “is it path-connected?” – that is – given any two points, is there a continuous function from the unit interval \([0,1]\) that starts on the first point and ends on the second.

Answering this question is, in general, a hard™ thing to do. An interesting although simple question in topology (which is a longwinded way to say “cute”) regards the path-connectivity of the real plane \(\mathbb R^2\) minus certain sets \(E\) . Namely, if \(E\) is countable, then \(\mathbb R^2\setminus E\) is in fact path connected.

This is somewhat surprising to me, still, considering that \(\mathbb Q^2\) is a subset of \(\mathbb R^2\) and there are no obvious paths connecting any two points. Let’s prove this result really quickly.

Take any two distinct points \(a\) and \(b\) on \(\mathbb R^2\setminus E\) . Since the latter is countable and the former isn’t, there are uncountably many points in that difference besides the two we have initially selected.

Moreover, there are uncountably many points not in \(E\) which are also not colinear with \(\overline{ab}\) . Let \(x\) and \(y\) be among those many points.

The line passing through \(x\) and \(y\) has uncountably many points on it; of which only countably many can be in \(E\) . So \(\overline {xy}\) really contains an uncountable quantity of points not in \(E\) .

Given any one such point \(p\) , we can trace segments \(\overline{ap}\) and \(\overline{pb}\) . These segments always have exactly three points in common, namely \(p\) itself, \(a\) and \(b\) : as \(a\) and \(b\) are distinct and \(x\) and \(y\) do not lie in \(\overline{ab}\) . The two segments together form a path from \(a\) to \(b\) which varies with \(p\) . Our goal is to show that there must be on such \(p\) for which the path we chose does not contain any points from \(E\) .

There are uncountably many such paths, one pair of segments for each \(p\) . For \(E\) to meet all paths, there would have to be a point \(e\) in \(E\) shared by uncountably many paths: for if that was not the case, then any point in \(E\)
is shared by at most a countable number of paths; call that set of paths \(\mathcal P(e)\) \[ \mathcal P(e) = \{ \overline{ap}\cup\overline{pb} \mid (p\in\overline{xy}\setminus E) \land (e \in \overline{ap}\cup\overline{pb}) \} \] Those families are all countable, and therefore the union of those countably many families is — itself — countable as well. Since there are uncountably many such paths, this couldn’t be the case.

Well, what’s the big deal about there being a point \(e\in E\) such that \(\mathcal P(e)\) is uncountable? I claim \(\mathcal P(e)\) always has at most one element. Think about it: a pair of paths \(\overline{ap}\cup\overline{pb}\) \(\overline{ap'}\cup\overline{p'b}\) have exactly three points in common: \(a\) , \(b\) and \(p\) . If \(e\) was in both, then it would have to be one of those three; but that’s impossible, as we know \(a,b,p\in\mathbb R^2\setminus E\) .

Which means that by assuming \(E\) could meet all paths we have arrived that there is an \(e\in E\) for which the \(\mathcal P(e)\) is big but at the same time, we know they have to be small due to just basic geometry. This means that the assumption of \(E\) meeting all paths is bogus, for it leads to a contradiction, and we can happily conclude that there is a path which \(E\) misses.

This path, then, is contained in \(\mathbb R^2 \setminus E\) and that proves path-connectedness.