# Real Analysis - Limit points and Open set.

The statement is If $$A \subset \mathbb{R}$$ is any set, $$I$$ is an open set that contains a limit point of $$L_{A}$$ - the set of all limit points of $$A$$, then $$I$$ also contains a point of $$A$$. I approach this by letting $$x$$ in A with $$x$$ is a limit point of A. So for every $$\epsilon >0$$ the deleted neighborhood $$V^{*}(a) \cap A$$ is nonempty. From this how do I show that $$x$$ is also in $$I$$.

• it is a point of $L_A$. – Dong Le Oct 31 '18 at 4:04

If $$I$$ is an open set containing a limit point of $$A$$, call it $$x$$ as you have. Then since $$I$$ is open it contains a ball around $$x$$, say $$(x-\epsilon,x+\epsilon)$$. Then since $$x$$ is a limit point of $$A$$ we know that as you've already pointed out $$((x-\epsilon,x+\epsilon))\cap A\not=\emptyset$$, so since $$((x-\epsilon,x+\epsilon))\subseteq I$$ it follows that $$((x-\epsilon,x+\epsilon))\cap A\subseteq I$$, hence $$I$$ contains an element of $$A$$.

• I forgot to mention that the deleted neighborhood is $V^{*}_{\epsilon}(x) = (x-\epsilon, x) \cup (x, x+\epsilon)$ which basically removes the element $x$ from the ball. – Dong Le Oct 31 '18 at 4:04
• Yes, I realized my error. It's been corrected to adjust for the fact that we don't know $x\in A$, merely that it is a limit point. – Melody Oct 31 '18 at 4:06
• I see! Thank you very much!! – Dong Le Oct 31 '18 at 4:07
• Actually, I'm gonna change it back. It doesn't matter if we consider the deleted neighborhood or not. All we need to know is the neighborhood of $x$ has a point in $A$, as the entire neighborhood is in $I$. – Melody Oct 31 '18 at 4:10

Let $$X$$ be any topological space and $$A\subset X.$$ As per the comments by the proposer, a point $$x\in X$$ is a limit point of $$A$$ iff every nbhd of $$x$$ contains a member of $$A.$$

Suppose $$y$$ is a limit point of the set $$L_A$$ of limit points of $$A$$. If $$U$$ is any nbhd of $$y$$ then there is an open $$V$$ with $$y\in V\subset U.$$

Now $$V$$ is a nbhd of $$y$$ so there exists $$x\in V\cap L_A.$$ But $$V$$ is also a nbhd of $$x$$ (because $$V$$ is open and $$x\in V$$ ), and $$x\in L_A$$, so there exists $$a\in A\cap V.$$

So $$a\in A\cap U$$ because $$V\subset U.$$

So any nbhd $$U$$ of $$y$$ contains a member of $$A.$$ So $$y\in L_A.$$

This holds in every topological space.