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Let $X$ be a metric space such that for every pair of disjoint open sets $U$ and $W$ we have $\overline{U} \cap \overline{W} = \emptyset$. How to prove $X$ is a discrete space?

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up vote 9 down vote accepted

The basic idea is to show that if there is a non-isolated point, there must be two disjoint open sets that both have that point as a limit point. One way to do that is to take a sequence converging to a non-isolated point and to put disjoint open sets around the odd-numbered and the even-numbered terms.

In more detail, suppose that $X$ is not discrete; then there’s some point $x\in X$ that isn’t isolated. Pick $x_0 \in N(x,1) \setminus \{x\}$, the open ball about $x$ of radius $1$ minus the point $x$, let $r_0 = d(x,x_0)/3$, where $d$ is the metric, and let $V_0 = N(x_0,r_0)$. Now pick $x_1 \in N(x,r_0)$, let $r_1 = d(x,x_1)/3$, and let $V_1 = N(x_1,r_1)$. At stage $n$ with $n>0$ pick $x_n \in N(x,r_{n-1})$, let $r_n = d(x,x_n)/3$, and let $V_n = N(x_n,r_n)$. You can check that the sequence $\langle x_n:n\in\omega\rangle$ converges to $x$, and the open sets $V_n$ are pairwise disjoint.

Let $U = \bigcup\limits_{n\in\omega} V_{2n}$ and $W = \bigcup\limits_{n\in\omega}V_{2n+1}$; $U$ and $W$ are disjoint open sets, so $\operatorname{cl}U \cap \operatorname{cl}W = \varnothing$. But $\langle x_{2n}:n\in\omega\rangle$ and $\langle x_{2n+1}:n\in\omega\rangle$ both converge to $x$, so $x\in \operatorname{cl}U \cap \operatorname{cl}W$, a contradiction.

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Excellent argument. And I think I can see how to generalise it to an arbitrary space $X$ usings nets. But here's a small nitpick: Technically you're not doing a proof by contradiction, you're proving the contraposed, i.e. $X$ metric and not discrete $\implies$ there exists disjoint open $U,V$ such that $\overline{U} \cap \overline{V} \neq \emptyset$. – kahen Aug 7 '11 at 21:35
As long as we're nitpicking: you mean to pick $x_0$ from $N(x, 1) \setminus \{x\}$, and so on. Great answer! – Dylan Moreland Aug 7 '11 at 21:39
@kahen: The proposition can not be generalised to arbitrary spaces since the hypothesis holds in any indiscrete space. – LostInMath Aug 7 '11 at 21:54
@Brian M. Scott: very very nice, thanks! – user10 Aug 7 '11 at 21:57
@LostInMath yeah... I figured that out myself a bit later. Too late to edit it now though. – kahen Aug 7 '11 at 22:00

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