2
$\begingroup$

$\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

$\endgroup$
  • 1
    $\begingroup$ I'm not familiar with the notation $X'$. Would you mind explaining what it means? $\endgroup$ – Robert Shore Feb 15 '19 at 0:09
  • 1
    $\begingroup$ Is the set of accumulation points of $X$. $\endgroup$ – Juan Daniel Valdivia Fuentes Feb 15 '19 at 0:10
  • $\begingroup$ It seems obvious that if $Y \subseteq X$, then any accumulation point of $Y$ must also be an accumulation point of $X$. $\endgroup$ – Robert Shore Feb 15 '19 at 0:14
2
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.