Does every uncountable subset of $R$ have a closed subset containing infinite elements? I am looking for a proof or counterexample. If the complement is countable it does have a perfect subset thats the best I could do.

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    $\begingroup$ Every subset of $\Bbb R$ with at least two elements has a closed subset with exactly two elements... $\endgroup$ – Lord Shark the Unknown Jun 10 '17 at 4:36
  • $\begingroup$ You meant Perfect ? right? $\endgroup$ – Red shoes Jun 10 '17 at 4:37
  • $\begingroup$ I mistyped prevoiusly...the perfect case is something I asked and got my answer...this is more general. $\endgroup$ – CoffeeCCD Jun 10 '17 at 4:48
  • $\begingroup$ This is pretty close to being a duplicate of math.stackexchange.com/questions/310113/… $\endgroup$ – Chris Culter Jun 10 '17 at 5:22

Every uncountable set $A\subset\mathbb{R}$ contains an accumulation point. I will prove this assertion. Suppose otherwise, that A contains none of its accumulation points. Then, for every element $a\in A$ there exists an $\epsilon_a$ such that the only element of $A$ in $(a-\epsilon_a,a+\epsilon_a)$ is $a$. Because of the density of $\mathbb{Q}$ in $\mathbb{R}$, there is a rational $q_a$ in $(a-\epsilon_a,a+\epsilon_a)$ for each $a\in A$. This defines a bijection between $A$ and a subset of $\mathbb{Q}$ and contradicts the assumption of $A$ being uncountable.

Now, suppose $x\in A$ is an accumulation point of $A$. Then, you can find a sequence $\{x_n\}_n\subset A$ such that $x_n\to x$. So, the set $\{x_1,x_2,\dots\}\cup\{x\}$ is closed and infinite.

  • $\begingroup$ Yes, so now you only need to prove the first sentence. $\endgroup$ – Andrés E. Caicedo Jun 10 '17 at 13:52

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