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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 $I$.

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

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    $\begingroup$ 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. $\endgroup$ – Dong Le Oct 31 '18 at 4:04
  • $\begingroup$ 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. $\endgroup$ – Melody Oct 31 '18 at 4:06
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    $\begingroup$ I see! Thank you very much!! $\endgroup$ – Dong Le Oct 31 '18 at 4:07
  • $\begingroup$ 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 $I$. $\endgroup$ – Melody Oct 31 '18 at 4:10
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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.

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