Let $A \subseteq \mathbb{R}^n$. Prove that if $A'$ is countable, then $A$ is countable.
$A'$ is the collection of all limit points of $A$.
This is the problem on Real Analysis. I know how to prove injection on $A$ to $\mathbb N=\{0,1,2,\ldots\}$, but I have no idea how to do the above problem.
On
The only points of $A$ that are not also points of $A'$ are isolated points. Therfore, $$ |A| \le |A'| + |I| $$ where $I$ is the set of isolated points of $A$. To show $A$ is countable, as $A'$ is known to be countable, by the above it suffices to show that there can only be countably many isolated points. In turn, it suffices to show there are only countably many isolated points in the ball of radius $n$ about the origin, $B(0, n)$, i.e. $I \cap B(0,n)$ is countable.
Around each point $x \in (I \cap B(0,n))$, there is some ball $B(x, \epsilon_x)$ which does not contain any other points in $I$. Taking each $\epsilon_x$ to be less than $1$, these balls are all disjoint and inside $B(0, n+1)$. So $$ \bigcup_{x \in I} B(0, \epsilon_x) \subset B(0,n+1). $$ Now, this is a disjoint union, and individual balls are Lebesgue-measurable, so (even if it is an uncountable sum!) $$ \sum_{x \in I} \mu(B(0, \epsilon_x)) \le \mu(B(0,n+1)) < \infty. $$ Each element of the sum is positive, and a sum of uncountably many positive real numbers is necessarily $\infty$. This would contradict the above, so $I$ must be countable.
HINT: Prove the contrapositive. That is, show that if $A$ is uncountable, then so is $A'$. One way to do this is to let $\mathscr{B}$ be a countable base for $\Bbb R^n$, let $\mathscr{B}_0=\{B\in\mathscr{B}:B\cap A\text{ is countable}\}$, and show that $A\setminus\bigcup\mathscr{B}_0$ is an uncountable subset of $A'$.