# Cl(A) smallest closed set containing A.

1. Sep 21, 2010

### Buri

My professor proved this result in class, but I don't understand the "simple" direction. He said that the above result is in another words proving that Cl(a) = intersection of all closed sets that contain A.

So he proved Cl(A) subset of intersection of all of the closed sets containing A. And intersection of closed sets containing A is a subset of Cl(A). However, I don't understand this last direction.

He simply says:

Cl(A) = int(A) U Bd(A); by definition.
=> Cl(A) is one of the closed sets containing A.
=> intersection of closed sets containing A is a subset of Cl(A).

That's what I just don't get. I understand that Cl(A) is closed and it contains A, but I just don't see how we can then say that the intersection of all closed sets containing A has Cl(A) in it. I know its going to be one of those I'll be intersecting all the other closed sets with, but that Cl(A) is actually INSIDE the intersection I just don't see.

I'm having a massive brain freeze so an explanation would be appreciated.

Thanks

2. Sep 21, 2010

### CompuChip

If you intersect two sets, say A and B, then the intersection will be a subset of A. After all, it contains only elements in A (which satisfy a specific property, namely that they are also elements in B).
By the same argument, it will also be a subset of B.

Now suppose that you intersect Cl(A) with I, the intersection of all other closed sets containing A. Then Cl(A) intersected with I is a subset of both Cl(A) and of I. It is the former you are interested in here.

3. Sep 21, 2010

### Buri

Ahhh thanks a lot. I get it. Stupid me just wasn't seeing it. Thanks a lot!