${ℝ}^2\setminus C$ is path connected if $C$ is countable

Theorem 1.

Let $C$ be a countable subset of ${\mathrm{R}}^{\mathrm{2}}$. Then ${\mathrm{R}}^{\mathrm{2}}\mathrm{\setminus }C$ is path connected.

We use ${ℝ}^2$ simply as an example; an analogous proof will work for any ${ℝ}^n,n>1$.

Proof.

Fix a point $P$ not in $C$. The strategy of the proof is to construct a path $p_x$ from any $x\in {ℝ}^2\setminus C$ to $P$. If we can do this then for any $d,d^{\prime }\in {ℝ}^2\setminus C$ we may construct a path from $d$ to $d^{\prime }$ by first following $p_d$ and then following $p_{d^{\prime }}$ in reverse.

Fix $x\in {ℝ}^2\setminus C$, and consider the set of all (straight) lines through $x$. There are uncountably many of these and they meet in the single point $x$, so not all of them contain a point of $C$. Choose one that doesn’t and move along it: your distance from $P$ takes on uncountably many values, and hence at some point this distance $r$ from $P$ is not shared by any point of $C$. The whole of the circle with radius $r$, centre $P$, lies in ${ℝ}^2\setminus C$ so we may move around it freely.

Consider all lines through $P$: these all intersect this circle, and there are uncountably many of them so we may choose one, say $L$, that contains no point of $C$. Moving around the circle until we meet $L$ and then following it inwards completes our path form $x$ to $P$.∎

Corollary 1.

Let $f\mathrm{:}{\mathrm{R}}^{\mathrm{2}}\mathrm{\to }\mathrm{R}$ be continuous and onto. Then $f^{\mathrm{-}\mathrm{1}}\mathit{}\mathrm{(}\mathrm{0}\mathrm{)}$ is uncountable.

Proof.

Suppose that $f^{-1}(0)$ is countable. ${ℝ}^2$ can be written as the disjoint union

$$f^{-1}(0)\cup f^{-1}((-\mathrm{\infty },0))\cup f^{-1}((0,\mathrm{\infty }))$$

where the last two sets are open (as $f$ is continuous), non-empty (as $f$ is onto) and disjoint. Since pathwise connected is the same as connected for Hausdorff spaces, we have that ${ℝ}^2\setminus f^{-1}(0)$ is not path connected, contradicting the theorem.∎