Solving a problem on metric spaces using measure theory

Question

I recently came across the theorem that every uncountable subset of $\mathbb{R}$ contains a limit point of that set. There is a straightforward proof, if we suppose that each point in some uncountable set $S$ is isolated, then by the density property of the rationals, we can associate each point in $S$ with a rational number in a unique way, which is a contradiction since $S$ is assumed to be uncountable.

I want to know if it possible to prove the following, weaker statement using measure theory, instead of the density property:

Every uncountable subset $S$ of $(0,1)$ contains a limit point of $S$

Intuitively, if we suppose all points of $S$ are isolated, then we can find some open set around each point of $S$, contained in $(0,1)$, such that the collection of such open sets, call it $\mathcal{G}$ has pairwise disjoint members, and then it seems like we should be able to conclude that the measure of $\bigcup\mathcal{G}> 1$, which would be the desired contradiction. But I don't have a formal background in measure theory, so I'm not really sure if this line of reasoning leads to anything. My motivation for this is that it would likely provide a generalisation of the above claim, to spaces where we don't know anything about dense subsets.

Answer 1

[Update: In the end, it turned out that both this answer and the previous answer by Didier were missing an important subtlety, as discussed in the comments. Since Didier has now addressed that subtlety, this answer became somewhat redundant, so I have modified it to discuss the matter in slightly more generality.]

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Claim. Let $X$ be a metric space, $\mu$ a finite Borel measure on $X$ for which open sets have positive measure. Then every discrete subset is countable.

Proof. If $S\subseteq X$ is discrete and uncountable, let $r_s=\frac{1}{2}d(s,S\backslash \{s\})$ for each $s\in S$. Then the balls $B_s:=B(s,r_s)$ are disjoint. Letting $S_n=\{s\in S\mid \mu(B_s)>\frac{1}{n}\}$, we see from discreteness that $S=\bigcup_n S_n$, so one of the sets $S_n$ is uncountable. Then by pairwise disjointness and finite additivity, we have $$\mu(X)\geq \sum_{s\in S_n}(\mu(B_s))\geq \sum_{s\in S_n} \frac{1}{n} = \infty\cdot,$$ contradicting finiteness of $\mu(X)$.

Remarks.

Of course, the conclusion of the claim above, in the context of metric spaces, is equivalent to the conclusion that $X$ must be separable/Lindelöf/Second countable.

Note that despite having an uncountable sum, only finite additivity and monotonicity of the measure is needed here.

Also, the essential ingredients for the above claim were:

1. In a general topological space with a finite Borel measure, open sets having positive measure immediately implies there can be at most countably many disjoint open subsets. This is known as a countable chain condition. It is equivalent in the metric space setting to separability, but in general is a slightly weaker property.

2. In a metric space $X$, for every discrete subset $S\subseteq X$ there is a family of disjoint open subsets covering $S$, with each subset containing exactly one member of $S$.

I do not know offhand to what extent the assumption of metrizability in 2 can be relaxed.

Answer 2

Lemma. Let $I$ be any set, and $\{a_i\}_{i\in I}$ be a set of non-negative real numbers indexed by $I$. If $\sum_{i\in I} a_i < + \infty$, then $\{i \in I \mid a_i \neq 0\}$ is at most countable.

Proof. Let $A_0 = \{i \in I \mid a_i \geqslant 1\}$, and for $n \geqslant 1$, consider $A_n = \{i \in I \mid \frac{1}{n} > a_i \geqslant \frac{1}{n+1}\}$. Since $\sum_{n = 0}^{\infty} \frac{|A_n|}{n+1} \leqslant \sum_{i\in I} a_i <+ \infty$, $A_n$ is finite for any $n \geqslant 0$. Note that $\{i \in I \mid a_i \neq 0\} = \cup_{n=0}^{\infty} A_n$ is a countable union of finite sets, and is thus at most countable. $\square$

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Assume $S \subset (0,1)$ is such that for each $s \in S$, there exists an open subset $U_s \subset (0,1)$ such that $S \cap U_s = \{s\}$, and such that $U_s \cap U_{s'} = \varnothing$ if $s \neq s'$. Let $\lambda$ be the Lebesgue measure. Since the $U_s$ are pairwise disjoint, we have, by this, that $$ \sum_{s\in S} \lambda(U_s) \leqslant \lambda(\cup_{s\in S}U_s) \leqslant \lambda((0,1)) = 1 < + \infty. $$ By the above Lemma, $S = \{s \in S \mid \lambda(U_s) \neq 0\}$ is at most countable.