# Basic point set topology question

1. Sep 8, 2008

### Diffy

As I understand it a topology on a set X is a collection of subset that satisfy three conditions

1) The collection contains X and the null set
2) It is closed under unions (perhaps a better way to say this is any union sets in this collection is again in the collection).
3) The intersection of any finite number of sets is in the collection.

Also I understand the Basis for a topology again to be a collection of subsets in X such that:

1) if x is in X than there is at least one basis element containing x.
2) if x is in any intersection of Basis elements there is a third basis element which contains the entire intersection.

So say I have a set X and I can form a basis for a topology B for X. Does that mean that X automatically has a topology? How do these two definitions relate? If there is a topology on a set does that meant there has to be a basis? Can someone help me out here and maybe explain things a bit?

2. Sep 8, 2008

### CompuChip

Well, the topology is often very large. For example, you can write down all topologies with 1, 2, 3 and 4 open sets but after that it becomes intractable. If you want to write down the standard topology on the real line, you are completely lost. A basis is sort of a small set generating the topology. Often, we specify a basis, and then work with the "smallest" topology in which all the elements of the basis are open sets (so we start from the basis and take all unions and intersections, repeating that process until we get nothing new).

For the example of the real line: a simple basis is the set { (a,b) | a, b real numbers } of open intervals. In fact you can take them to be rational, and it turns out that the basis is countable! If you want to prove something about the uncountable collection of all open sets, it suffices to prove it for the countable basis. Any open set is just a union or finite intersection of basis elements.

There is a theorem, that every basis generates one (and only one) topology. On the other hand, every topology has at least one basis: the topology itself But often a smaller basis exists.
But as I already pointed out, we usually use the basis to construct the topology. We start out with a set with some intuitive ideas of which sets we would like to be open. The concept of basis is a good idea to find a complete topology, in which at least those sets are indeed open. In the case of the real line: we would like the intuitive things we call "open intervals" to be open in the formal sense (i.e. elements of the topology), so we start with a basis of those. From there we can construct all other open sets (e.g. unions of open intervals).

3. Sep 8, 2008

### Diffy

Great answer, and thanks, that gives me a good understanding of the necessity and use of the definition I have for a basis.

I will follow up by just proving that the set {(a,b) | a, b real numbers} is indeed a basis of the real line. Let me know if what I write is any good.

Let x be a real number, then (x - n, x + n) | n is a Positive Real number, is in {(a,b) | a, b real numbers} and contains x.

Let S = (a, b) intersect (c, d); then (Min(a, c), Max(b, d)) is both in {(a,b) | a, b real numbers} and contains S.

So assuming I did it right, I've proved the set {(a,b) | a, b real numbers} is a basis for a topology on the real line. I see that it is not yet a topology because it doesn't contain the Null set. So how do I go about formally, or can I simply state that the topology on the real line can be formed by taking all unions and finite intersections of my set {(a,b) | a, b real numbers}?

Hope this makes sense, topology isn't my cup of tea.

4. Sep 8, 2008

### gel

mmm...doughnuts.

5. Sep 8, 2008

### gel

Yes, so the set of all unions of intervals of the form (a,b) is a topology on the real line.

6. Sep 9, 2008

### CompuChip

I only now noticed that your point 2 differs from what I learned (I don't know if it is wrong or simply one of those many equivalent definitions in topology):

2) If B1, B2 are basis elements, then for any x in the intersection there is another basis element B which is completely contained in the intersection which contains x.

I always think of this as the formalization of open balls in Euclidean space: if you have the intersection of two open balls, every point is an interior point (i.e. there is an open ball around it completely contained inside the intersection).

So for the definition given you did it right, but I doubt the definition
You can indeed make a topology out of it by taking unions and finite intersections. For example, the union of all basis elements gives all of the space, the intersection of any disjoint intervals gives the null set, etc.

Assuming you mean, $S^1 \times S^1$ ?
By the way, it's the same as a cup of tea. At least, they are homotopically equivalent.

7. Sep 9, 2008

8. Sep 9, 2008

### CompuChip

That's always good to hear

It leaves you to still prove the second property though