# Removing uncountably many points from an open set leaving it open

Can we remove uncountably many points from $\mathbb R^n$, without removing all points from any open ball, to leave $\mathbb R^n$ open (with the usual topology)?

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For example when $n=2$ you can remove $\{(x,y),x=0\}$ (if I didn't misunderstand the question). – Davide Giraudo Apr 7 '12 at 13:04

## 3 Answers

Davide's example easily generalises to all n>1. For n=1, I believe removing the Cantor set would work.

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Removing the Cantor set certainly works. – Carl Mummert Apr 7 '12 at 13:25

Order the points with rational coordinates as the sequence $(q_k)_k$ and consider the union $U$ of the open balls centered at $q_k$ with radius $r_k$, for some positive sequence $(r_k)_k$. The volume of $U$ is less than a multiple of $s=\sum\limits_k(r_k)^n$. Assume $s$ is finite.

Then the Lebesgue measure of the complement $V$ of $U$ is nonzero (in fact it is infinite) hence $V$ is uncountable. Furthermore, $U$ is open and any open ball contains at least one point $q_k$, hence it meets $U$.

This construction works for every $n\geqslant1$.

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Take $S:=\{(0,0,\ldots,0,a),a\in\mathbb R\}\subset \mathbb R^n$. Then $S$ is uncountable since it's in bijection with $\mathbb R$ and doesn't contain any ball (if it contains the ball $B((x_1,\ldots,x_n),r)$ then $x\in S$ so $x_j=0$ for $1\leq j\leq n-1$ and $\left(\frac r2,0,\ldots,0,x_n\right)\in B((x_1,\ldots,x_n),r)\subset S$, a contradiction.

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