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Let $A$ be an uncountable set of real numbers. Then by using bolzano-weierstrass theorem, it can be easily shown that $A$ has at least one limit point. But my question is how many limit point? Can an uncountable set has countable number of limit points? Please suggest me. Thanks in advance.

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  • $\begingroup$ Let $U = \bigl\{ x \in \mathbb{R} : \bigl(\exists \varepsilon > 0\bigr)\bigl(A \cap (x-\varepsilon, x+\varepsilon) \text{ is countable}\bigr)\bigr\}$. What do you know about $U$? $\endgroup$ – Daniel Fischer Sep 21 '15 at 11:23
  • $\begingroup$ then what you have shown? $\endgroup$ – neelkanth Sep 21 '15 at 11:25
  • $\begingroup$ @neela: See this answer. $\endgroup$ – Brian M. Scott Sep 21 '15 at 11:32
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Without loss of generality, we can assume $A$ is closed, since the limit points of $\overline{A}$ are exactly the limit points of $A$. The Cantor-Bendixson Theorem says that every uncountable, closed subset of the reals contains a nonempty perfect subset. (A perfect set is a set with no isolated points, i.e., a set where every point is a limit point.) It is a fact that every (nonempty) perfect subset of $\mathbb{R}$ has cardinality $2^{\aleph_0}$ (proof idea: build a tree of closed sets, and use Bolzanno-Weierstrass to show that every path along the tree contains a distinct point), $A$ has at least $2^{\aleph_0}$ limit points. Since there are only $2^{\aleph_0}$ real numbers to begin with, this answer is exact.

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  • $\begingroup$ how assuming closed do not loss generality? $\endgroup$ – neelkanth Sep 23 '15 at 7:33
  • $\begingroup$ @neela In the more general case, the proof I wrote can show that $\overline{A}$ has $2^{\aleph_0}$ limit points. Since those limit points will all also be limit points of $A$, $A$ has $2^{\aleph_0}$ limit points, too. $\endgroup$ – Mike Haskel Sep 23 '15 at 10:14

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