# Closure boundary interior sets

If we denote for a set $A$: $A^{o}$ the interior points set; $\overline A$ the closure and $\delta A$ the boundary set and $A'$ the set of cluster points , do the following hold (give counter-examples where not)?

• $A^{'} \subseteq A^{''}$ or $A^{''}\subseteq A^{'}$

I cannot come up with some sets on $R$ to check these relations so I would really appreciate some help. Thank you!

• It might be better to concentrate on a couple of these, rather than ask essentially 8 different questions in one post. Once you see how to do one of them, you can start to figure out how to do the others. Oct 28 '13 at 15:57
• I actually did all of them but this one with the cluster points. Oct 28 '13 at 16:01
• Ah, okay. I will edit my answer. Oct 28 '13 at 16:01

1. If $A = \{1/n : n\in \mathbb{N}\}$, then $A' = \{0\}$ and $A'' = \emptyset$, so $A'$ is not a subset of $A''$.
2. If $x \in A''$, then there is a sequence $(x_n) \in A'$ such that $x_n \to x$. Now for any $\epsilon > 0$, there is $y_n \in A$ such that $$|x_n - y_n| < \epsilon/2$$ and $N_0 \in \mathbb{N}$ such that $$|x_n - x| < \epsilon/2 \quad\forall n\geq N_0$$ Hence, for any $n \geq N_0$, $$|y_n - x| < \epsilon$$ Hence $y_n \to x$, and so $x \in A'$. Hence $$A'' \subset A'$$
Alternatively, if $x\in A''$, we want to show that $x\in A'$ :
If $U$ is an open neighbourhood of $x$, then there exists $y \in A'\cap U$, with $y\neq x$. Choose a neighbourhood $V$ of $y$ such that $y \in V, x\notin V$, and $V\subset U$. Then there exists $z\in A\cap V$. Hence, $$z\in A\cap U, z\neq x$$ and hence $x\in A'$
First, I want to give an example where $A''\neq A'$. Consider the set $A=\{ (1/n,1/m)\,|\, n,m\in \mathbb Z_+\}\subset \mathbb R^2$.Then $A'=\{(0,1/m)\}\cup\{(1/n,0)\}$, so that $A''=\{(0,0)\}$.
Second, I will show that if we assume points are closed, then $A''\subseteq A'$. Suppose $p$ is a cluster point of $A'$. Then every neighborhood $U$ of $p$ intersects $A'$ in a point $q$ other than $p$. Hence $U\setminus\{p\}$ is a neighborhood of $q$, and must intersect $A$ in some point. Thus $U$ also meets $A$ in a point other than $p$, and so $p$ is a cluster point of $A$.