2
$\begingroup$

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!

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Oct 28 '13 at 15:57
  • $\begingroup$ I actually did all of them but this one with the cluster points. $\endgroup$ Oct 28 '13 at 16:01
  • $\begingroup$ Ah, okay. I will edit my answer. $\endgroup$ Oct 28 '13 at 16:01
2
$\begingroup$
  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 $x\in A'$

$\endgroup$
1
$\begingroup$

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 $A$.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.