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If I have an uncountable subset $A \in \mathbb{R}$, and we assume A is nonempty, does it follow that every point within $A$ is a limit point of $A$ from the density of $\mathbb{Q}$ in $\mathbb{R}$ (i.e. between any two real numbers there exists one rational number, and by extension infinitely many rationals)?

As far as I can tell, this logic would not hold for all countable subsets, since I believe, for example, that $\mathbb{Z}$ has no limit points whatsoever.

I'm trying to straighten out the concept of limit points in my head, and I'd love some clarification of my logic. Many thanks!

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    @Trevor: Probably...2012-10-12

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As Chris has pointed out, the answer is no. I would like to elaborate on this a bit.

A closed set whose points are all limit points of the set (i.e., a closed set with no isolated points) is called a perfect set. Every perfect set in $\mathbb{R}$ is uncountable and in fact contains a homeomorphic copy of the Cantor set. There are plenty of uncountable sets that are not perfect, such as the one in Chris's answer. In fact with the Axiom of Choice one can show that there are uncountable sets that do not even have perfect subsets (see http://en.wikipedia.org/wiki/Bernstein_set,) providing a strong sort of counterexample to the original question.

It is true, however, that every uncountable closed set contains a perfect subset. This is the Cantor–Bendixson theorem (see http://en.wikipedia.org/wiki/Perfect_set_property.) For example, the closed set $\{0\} \cup [1,2]$ in Chris's answer, while not perfect, contains the perfect subset $[1,2]$.

Finally, it is consistent with ZF (without the Axiom of Choice) that every uncountable set (closed or not) contains a perfect subset. (See http://en.wikipedia.org/wiki/Perfect_set_property.)

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    @CarlMummert Thanks. I have corrected the definition and expanded my answer.2012-10-12
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No, of course not. I don't understand your reasoning, but the conclusion is wrong. For example, $\{0\} \cup [1,2]$ is uncountable, but $0$ is not a limit point.

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    So what? You claim that *every* point of $A$ is a limit point: I have exhibited one which is not. The status of the other points of $A$ is irrelevant.2012-10-09