[Math] Proving the set of all cluster points is closed

general-topologyreal-analysis

if $A$ is a subset of $\mathbb{R}^n$. Can I prove that "the set of all cluster points of A is closed" using the following:

  • A closed subset is a subset that contains all its limit points.
  • Every cluster point is a limit point (Not vice versa though).
  • A has all cluster points means that it includes all its limits point and hence
    A is closed.

Is this an enough proof ?

Best Answer

hint. Let $C$ be the set of cluster points of $A$. First you need to state what you need to prove. You need to prove that if $c$ is a limit point of $C$, then $c\in C$. In other words, you need to prove that if $c$ is a limit point of $C$ then $c$ is a cluster points of $A$. Since you work with subsets of $\mathbb R^n$, it is enough to work with sequences. You may pick a sequence $c_n,$ $n\ge1,$ of points of $C$ with $\lim_{n\to\infty}c_n=c.$ For each $n$ you may pick a sequence of points $a_{n,m}$, $m\ge1$, in $A$, such that $\lim_{m\to\infty}a_{n,m}=c_n.$ Now, you could either consider an arbitrary neighborhood $O$ of $c$ and show that $O$ necessarily intersects $A$, or, alternatively, you could pick suitable $m(n)$ for each $n$ such that $\lim_{n\to\infty}a_{n,m(n)}=c.$

Say, for each $n$ pick $m(n)$ big enough such that the distance from $a_{n,m(n)}$ to $c_n$ is less than $\frac1n.$ Then show that necessarily $\lim_{n\to\infty}a_{n,m(n)}=c.$ (If you are not sure if my hint is complete, you may wish to consider more cases, depending of the distinction between cluster points and limit points, and keep track which points may belong to which set. But the main case would be similar to the hint I sketched here.)

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