Thread: Topologic spaces question...

Hello, thanks for reading.
Let (X,T) be a topologic space.
I got the next definition of "the closure of A, partial to X":
cl(A) = intersection (F | A is partial to F, F is closed in (X,T) )
I need to prove that cl(A) = AUA', where A' is the set of all accumulation points of A.

And aI don't succeed :-\

Thanks!
Tomer

Originally Posted by aurora

Let (X,T) be a topologic space.
I got the next definition of "the closure of A, partial to X":
cl(A) = intersection (F | A is partial to F, F is closed in (X,T) )
I need to prove that cl(A) = AUA', where A' is the set of all accumulation points of A.

Not sure what term you are translating as “partial” but the English term is “subset” in this context.

Now it should be clear that . If then is a limit point of so if then .
But that means that and the complement is open.
What is the contradiction there?
Can you finish?

I finished it, I think , thank you very much!
Sorry for the "partial" thing... How do you call that symbol of subsets in English?
For the other direction, I wrote something, but something about it looks kinda akward, I'd really appreciate it if you check it out for me.
Before that, however - would you be willing to explain to me how to use math symbols in here, or reffer me to instructions? I couldn't find them. I don't want to wear you out with my "partial" English

Again, thank you!

Tomer.

This is a perfect example of how your text/instructor determines how statements are proved.
I think that you should have proved that the closure of a set is a closed set.
Therefore, one direction is trivial.

As for using mathematical symbols here, you need to learn to use LaTeX at a very basic level. There is a forum here on LaTeX help with suggestions on free programs that you can use.

Guess it's too trivial for me to grasp .
I've spent too much ink on this proof, apperently... Mind you explain why it's trivial?

It seems to me that your text is using the definition of closure.
Can you prove that that set is a closed itself?
If so then by definition .

If you have trouble proving is closed, think that any point not in that set is neither a point of A nor a limit point of A. Therefore its complement is an open set so the set is closed.

Actually, my "text's" ( = lecturer) definition for a closed set is a set of which it's... complement (?) set is open.
erm... ok. A is closed if "X-A" (or A^c) is open.