I can´t prove this fact in $\mathbb{R}$. I want to know how general this result is. (What topological properties are needed to prove it?) Let $X$ be a non-denumerable subset of the real numbers. How can I prove that the set of limit points of $X$ is also non-denumerable?
A non-denumerable set has non-denumerably many cluster points?
0
$\begingroup$
real-analysis
general-topology
-
0I believe the common term is non-denumerable. – 2011-08-23
1 Answers
5
Let Y be the set of limit points of X.
Since X-Y does not contain any of its limit points, X-Y is discrete.
We need that all discrete subsets are (at-most-)countable.
With that, since $X\subseteq (X-Y)\cup Y$, $\; $ Y is not (at-most-)countable.
A discrete subset of a separable metric space is always (at-most-)countable.
-
0Beautiful, and thank! – 2011-08-24