# Topologic spaces question...

• Jun 6th 2008, 02:50 PM
aurora
Topologic spaces question...
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 :)
• Jun 6th 2008, 03:12 PM
Plato
Quote:

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 $A \subseteq \cap \left\{ {F:F \mbox { is closed } \wedge A \subseteq F} \right\}
$
. If $p \in A'$ then $p$ is a limit point of $A$ so if $p \notin \cap \left\{ {F:F\mbox { is closed } \wedge A \subseteq F} \right\}$ then $\exists E \in \left\{ {F:F\mbox { is closed } \wedge A \subseteq F} \right\} \wedge p \notin E$.
But that means that $p \in E^c$ and the complement $E^c$ is open.
Can you finish?
• Jun 7th 2008, 02:02 AM
aurora
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 (Rofl)

Again, thank you!

Tomer.
• Jun 7th 2008, 04:06 AM
Plato
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.
• Jun 7th 2008, 09:33 AM
aurora
Guess it's too trivial for me to grasp (Crying).
I've spent too much ink on this proof, apperently... Mind you explain why it's trivial?
• Jun 7th 2008, 11:15 AM
Plato
It seems to me that your text is using $A \cup A'$ the definition of closure.
Can you prove that that set is a closed itself?
If so then by definition $A \subseteq A \cup A'\quad \Rightarrow \quad A \cup A' \in \left\{ {F:F\mbox{ is closed } \wedge A \subseteq F} \right\}$.

If you have trouble proving $A \cup A'$ 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.
• Jun 8th 2008, 12:55 AM
aurora
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.