# Topology Question

1. Sep 27, 2005

### philosophking

Hey everyone,

First of all, I hope this is an OK place for a topology question. I was debating here or of course set theory, but I guess this is right. Anyway.

I'm studying out of Munkres' book, and I'm looking at a certain problem. The problem stated is:

THEOREM: If A is a basis for a topology on X, then the topology generated by A equals the intersection of all topologies on X that contain A.

Basically, I interpreted this to mean that the basis for a topology on X is necessarily an element of every topology on X, and here's my proof:

PROOF: Suppose to the contrary, that there exists an element in the intersection of all topologies that is not in A. Hence there exists an open set U in X such that for each a in U, there is no basis element A_i such that x is in A_i and A_i is a subset of U. But since this is a topology, x must be in the intersection of U and some element of our basis collection. This would create a basis for x which is a contradiction.

See, I'm really not sure if I did this correctly. My "leap" is where I assume the open set containing U intersects A_i, I think, and that's where I think I go wrong. Could anyone help? Thanks in advance!

2. Sep 27, 2005

### Hurkyl

Staff Emeritus
Well, I'll correct your mistakes, and give you a chance to work through it again:

"the intersection of all topologies on X that contain A."

There are (generally) topologies that do not contain A. (e.g. the indiscrete topology)

Furthermore, a basis is not an element of a topology, it is a subset of a topology.

3. Sep 27, 2005

### AKG

Well the topology generated by A is just the collection of all unions of elements of A. Any topology containing A must contain all unions of elements of A by definition of a topology. So every topology containing A contains the topology generated by A. Therefore the intersection of all these topologies must contain the topology generated by A. It remains to show that the intersection contains nothing more. Well the intersection is contained in each topology being intersected, and the topology generated by A is one such topology since it is obviously a topology containing A. So the intersection must be contained in the topology generated by A, and hence the intersection is the topology generated by A.

If you're using Munkres' "Topology" 2nd Edition, the above makes use of Lemma 13.1. I find it hard to follow your proof. For one, A is just a basis so even the topology generated by A generally contains elements that aren't in A. Perhaps in the first line of your proof, you mean "the topology generated by A" and not just "A" but I'm not sure, which is why I say your proof is hard to follow. Another problem is that you say that there exists and open set U in X such that... but how can you say U is open without saying that it is open with respect to a given topology? Then you say, "for each a in U" but then never mention "a" again, but suddenly start mentioning "x". Maybe you meant "x" instead of "a" the first time? And what's the justification for your second sentence? I also don't follow the rest of your proof, but that may just be me.

4. Sep 28, 2005

### philosophking

Wow... I'm really sorry you guys. I almost feel embarrased that I put up that proof now. If you guys still want to help me out, thanks a lot. I used your suggestions and went back and redid the proof. It's still proving by contradiction. Here it is:

1. We want to show that either the intersection of all topologies containing A MINUS the topology generated by A does not equal the empty set OR

2. We want to show the topology generated by A MINUS the intersection of all topologies containing A does not equal the empty set.

We show 1.

Hence, There exists a set B in the intersection of all topologies such that B is in each topology CONTAINING A but not in the topology generated by A.

Then B is an open set in each of these topologies. Let A_r be an element in A. Since each set is a topology, B (intersect) A_r is in each topology. Furthermore, since A_r is a basis element, there exists an A' such that x (which is some element of B) is in A' and A' is a subset of A_r (intersect) B for each r in the index set of elements in A. But this would imply A' is in the topology generated by A, which is a contradiction (we are assuming B is a set outside of the basis set).