# Given a non-countable subset $X$ of $\mathbb{R}$, prove that $X'$ is uncountable

$$\textbf{My solution} :$$

We will use this theorem: If a set $$X$$ only has isolated points then $$X$$ is countable or equivalently if $$X$$ is uncountable then $$X$$ contains some point of accumulation in $$X$$ ($$X \cap X'\neq \emptyset$$)

Suppose that $$X'$$ is countable, then $$Y=X-X'$$ is uncountable (if it were we would have X is countable) then by the theorem we would have that there is some $$a \in Y'\cap Y$$.

We affirm that $$a$$ is point of accumulation of $$X$$.

Indeed, for any $$\epsilon >0 : (a- \epsilon,a+ \epsilon) \cap Y-\{a\} \neq \emptyset$$, then exists $$x_1 \in (a- \epsilon,a+ \epsilon) \cap Y-\{a\}$$. As $$x_1$$ is in $$Y$$ then $$x_1$$ is in $$X$$ so $$x_1 \in (a- \epsilon,a+ \epsilon) \cap X-\{a\}$$ then $$a \in X'$$ a contradiction because $$a \in Y$$.

I wanted to know if my solution is correct, something to improve or maybe some other. Thank you

• I'm not familiar with the notation $X'$. Would you mind explaining what it means? – Robert Shore Feb 15 '19 at 0:09
• Is the set of accumulation points of $X$. – Juan Daniel Valdivia Fuentes Feb 15 '19 at 0:10
• It seems obvious that if $Y \subseteq X$, then any accumulation point of $Y$ must also be an accumulation point of $X$. – Robert Shore Feb 15 '19 at 0:14

Suppose $$X$$ is uncountable. Let $$c(X)$$ be the set of condensation points of $$X$$: $$\{x \in X: \forall r>0: (x-r,x+r) \cap X \text{ uncountable}\}$$. Let $$B_n, n \in \mathbb{N}$$ be a countable base for $$\mathbb{R}$$ (e.g. all rational intervals).
All condensation points are trivially in $$X' \cap X$$. For every point $$p \in X\setminus c(X)$$ there is some $$n(p) \in \mathbb{N}$$ such that the neighbourhood $$B_{n(p)} \cap X$$ of $$p$$ is at most countable. Then $$X \setminus c(X)=\bigcup\{B_{n(p)}: p \in X\setminus c(X)\}$$ is a countable union of countable open sets (of $$X$$) and so as $$X$$ is uncountable we have that $$c(X)$$ is closed in $$X$$ and uncountable. So in particular $$X'$$ is uncountable (and non-empty, which is all you needed to prove).