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Let $X\subset\mathbb R$ and denote by $X'$ the set of limit points of $X$.

Say if it is true or false that:

If $X'$ is finite or countable, then $X$ is finite or coutable.

I know that if $X$ is finite or countable , then $X'$ is empty, then finite. But i get stuck there. Thank you

Joe
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  • If $X$ is finite then $X'$ is empty, but if $X$ is countable then $X'$ maybe nonempty. Say $X:={\frac 1 n:n\in \mathbb N}$ – Tongou Yang Jan 31 '17 at 16:04
  • Somewhat related: http://math.stackexchange.com/questions/468746/uncountable-set-with-exactly-one-limit-point – Henricus V. Jan 31 '17 at 16:05

1 Answers1

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Any uncountable subset $A$ of $\mathbb{R}$ has uncountably many limit points. This is true in any separable metric space. Take $B_n$ a countable open base for the topology (for the reals we can take all open intervals with rational endpoints), and suppose $A$ is uncountable and consider points not in $A'$:

For every $x \in A\setminus A'$, pick $B_n(x)$ such that $B_n(x) \cap A \subset \{x\}$, this can be done as $x$ is not a limit point of $A$. Clearly $x \neq y$ implies $n(x) \neq n(y)$.

So we have an injection from the set $A \setminus A'$ into the countable set of indices $\mathbb{N}$. So there are only countably many points of $A$ that are not limit points of $A$, and as $A$ is uncountable, $A'$ must be uncountable.

So indeed a set with at most countably many limit points is at most countable as well.

Henno Brandsma
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