0

Good afternoon. I Have a problem with this proof:

Let $B$ the set of limit points of $A$. Prove $B$ is a closed set.

I try to make this:

Suppose $B$ is not a closed set. Then exist a $B ∩B(x;r)=Ø$ where $r>0$. Suppose $A ⊆ B$ and $x$ a limit point of $A$ Then $x$ is a limit point of $B$. Contradiction.

Please, help me, i don't know if this is fine.

rcoder
  • 4,545
  • Not closed does not imply open in general! For example, $A = { \tfrac{1}{n} : n \in \mathbb N }$ is not closed in $\mathbb R$ (since $0$ is a limit point of $A$ not contained in $A$), and $A$ is not open, either, since for any point $1/n$ in $A$ we may pick a sufficiently small ball $B$ around $1/n$ so that $B \cap A = {1/n}$, and hence $A$ is not open. – Jon Warneke Jul 31 '16 at 18:57

1 Answers1

0

According to my basic knowledge that it's one of the definitions of closed set.

But if you demonstrate this statement with only to say $A^{c}$ it's closed if $A$ it's open. You can see it so.

If $A$ (the set with its limit points) is closed then $A^{c}$ it's open. So we'll show it.

Any point in $A^{c}$ has a neighbourhood completely inside of $A^{c}$. Why? As I see it. If you find any point in $A^{c}$ which you couldn't choose a neighbourhood which is contained completely in $A^{c}$ this point it's a limit point given any neighbourhood has as many elements of $A$ as of its complement. This means that this point belong to $A$ and by definition this doesn't belong to it's complement. With this has shown whatever point in $A^{c}$ has a neighbourhood completely inside of $A^{c}$ then It's an open set and again by definition its complement it's closed.

7919
  • 161