Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $E$ be a set in a metric space $X$ and $E'$ denote the set of all limit points and $\bar{E}$ be the closure of $E$. Then $\bar{E} = E \cup E'$

Prove that $E'$ is closed, and $\bar{E}$ and $E$ have the same limit points

So as sets, they look like?

enter image description here

I know that the $p'$ just solely inside $E'$ is wrong, but is the actual picture even right?

I read a lemma that since $E'$ is the set of all limit points, any point (say) $p'$ must contain another point in $E$ for that point isn't the point itself

share|improve this question
    
In $\Bbb R$, what are the limit points of $E=\{1/n: n=1,2,\ldots\}$? –  David Mitra Dec 29 '12 at 3:08
    
Is it just the unit interval? –  sidht Dec 29 '12 at 3:22
    
It's $0$. But to show that $E'$ is closed you can show that every convergent sequence in $E'$ has its limit in $E'$. –  André Dec 29 '12 at 3:24
    
Oh sorry I misread his question. I thought he asked what is $E'$ for some reason. I actually have the answers in my hand, and I need to understand the definitions first before I want to continue however –  sidht Dec 29 '12 at 3:32
    
Alternative hint: to show a set is closed, it is often easier to show its complement is open, –  gnometorule Dec 29 '12 at 3:54

1 Answer 1

up vote 1 down vote accepted

To show that $E'$ is closed, we will show $X\setminus E'$ is open. If $x \in X\setminus E'$, then there is $B(x,r) \subseteq X\setminus E$. Then for all $y \in B(x,r)$ since $B(x,r)$ is open there is $B(y,r_2)\subseteq B(x,r)\subseteq X \setminus E$ so $y \in X \setminus E'$ as well. It follows that $x$ is an interior point of $X\setminus E'$, and so $X\setminus E'$ is open, and $E'$ is closed.

Now we show $E$ and $\overline{E}$ have the same limit points. Clearly $E' \subseteq \overline{E}'$ since if every $B(x,r)$ contains an element of $E$ different from $x$, then that element is also an element of $\overline{E}$. Conversely, suppose $B(x,r)$ contains an element $y$ of $\overline{E}$. Since $B(x,r)$ is open, we may choose $B(y,r_2)\subseteq B(x,r)$ and by making $r_2<d(x,y)$ smaller if necessary we may ensure $x\notin B(y,r_2)$. Then since $y \in \overline{E}$ we may choose $z \in E \cap (B(y,r_2)\setminus\{y\})$. Then $z \in E \cap (B(x,r) \setminus \{x\})$ which completes the proof.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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