1. ## Closed set

If a subset of a top. space is closed, then it contains all its limit points.

So I need to show that if $\displaystyle A\subset X$ is closed then $\displaystyle A=\bar{A}$.

Since A is closed, $\displaystyle X-A$ is open. Also, $\displaystyle \bar{A}=A\cup A'$

A' is the set of all limit points of A.

I am not sure if this helps, but

$\displaystyle X-\bar{A}=X-(A\cup A')=(X-A)\cup (X-A')$

2. ## Re: Closed set

Originally Posted by dwsmith
If a subset of a top. space is closed, then it contains all its limit points.

So I need to show that if $\displaystyle A\subset X$ is closed then $\displaystyle A=\bar{A}$.

Since A is closed, $\displaystyle X-A$ is open. Also, $\displaystyle \bar{A}=A\cup A'$

A' is the set of all limit points of A.

I am not sure if this helps, but

$\displaystyle X-\bar{A}=X-(A\cup A')=(X-A)\cup (X-A')$
I'm confused, you're trying to prove that if $\displaystyle A$ is closed then $\displaystyle \overline{A}=A$ where $\displaystyle \overline{A}$ is DEFINED to be $\displaystyle A\cup A'$? Well, isn't $\displaystyle A'\subseteq A$ so that trivially $\displaystyle A\cup A'=A$?

3. ## Re: Closed set

Originally Posted by Drexel28
I'm confused, you're trying to prove that if $\displaystyle A$ is closed then $\displaystyle \overline{A}=A$ where $\displaystyle \overline{A}$ is DEFINED to be $\displaystyle A\cup A'$? Well, isn't $\displaystyle A'\subseteq A$ so that trivially $\displaystyle A\cup A'=A$?
No. The limit points don't have to to be in A.

4. ## Re: Closed set

Originally Posted by dwsmith
The limit points don't have to to be in A.
They must belong to A if A is closed.
In the OP you wrote: if $\displaystyle A$ is closed then $\displaystyle A=\overline{A}$.

Note that one way of defining $\displaystyle \overline{A}$ as the smallest closed set that contains $\displaystyle A$.

5. ## Re: Closed set

I think there is an issue. The also part is clarifying notation not part of the statement to prove.

6. ## Re: Closed set

Originally Posted by dwsmith
I think there is an issue. The also part is clarifying notation not part of the statement to prove.
A\subset X[/TEX] is closed then $\displaystyle A=\bar{A}$. Since A is closed,
Because $\displaystyle X\setminus A$ is open then by definition of limit point, all $\displaystyle A'\subseteq A$.