Help verifying my proof

Question

I needed to prove by contradiction that a countable set is convergent to its accumulation points.

Answer 1

Here is one way to do it: Let $I$ be the set of bounded open intervals with rational endpoints and let $J=\{i\in I: |S\cap i|\leq \aleph_0\}.$ Let $K=\bigcup J.$

Then $S\cap K=\cup_{j\in J}(S\cap j)$ is a countable union of countable sets, so $S\cap K$ is countable. So $S^*=S\setminus K=S\setminus (S\cap K)$ is an uncountable subset of $S.$

For any $x\in S^*$ and any $r>0,$ take $q,q'\in \Bbb Q$ with $-r+x<q<x<q'<r+x.$ Now if $S\cap (-r+x,r+x)$ were countable, then $S\cap (q,q')$ would be countable, implying $x\in (q,q')\in J,$ implying $x\in K,$ contrary to $x\in S^*=S\setminus K.$

So $S^*$ is an uncountable set such that every nbhd of every member of $S^*$ has uncountable intersection with $S.$

Furthermore, if $x\in S^*$ and $U$ is a nbhd of $x$ then $(S^*\cap U)\cup ((S\cap K)\cap U)=S\cap U$ is uncountable but $(S\cap K)\cap U$ is a subset of the countable set $S\cap K,$ so $S^*\cap U$ must be uncountable.