Some introductory Topology questions

jgens
Gold Member
So, is the typo that the second to last T should actually be an l (the sub collection)?
Yes.

Can a topology whose finite intersections form a basis for itself exist?
Yes. This is true for every topology.

Yes.

Yes. This is true for every topology.
Ohhh, that's even worse then, hah!

Thanks, I think I see why what you say is true now.

WannabeNewton
Wow, this book is plagued with errors.
jgens answered your questions so that takes care of that. I'm just curious as to what book this is?

jgens answered your main question so that takes care of that. I'm just curious as to what book this is?
"Foundations of Topology" C. Wayne Patty, 2nd Edition

But so I understand why what jgens said is true, we know that any intersection of members of T is also in T. And any topology is a basis for itself. So, the collection of all intersections of members of T is T, and thus is a basis on T.

Is that right?

WannabeNewton
Essentially yes.

If the book is marred with errors, you might want to use a more standard text.

Essentially yes.

If the book is marred with errors, you might want to use a more standard text.
I see what you mean. This is just the textbook that the class uses. Do you have any recommendations?

Thanks again.

A good book "Introduction to Topological Manifolds" by Lee. Don't let the title mislead you, it's more about topology than about the manifolds.

Then there is also Munkres, which is widely used.

But both books have their flaws. For example, infinite products and the Tychonoff theorem aren't treated well in either book. And nets and filters aren't treated in the books either, or at least not very well.

A better book is Willard, but it is also quite advanced (and has some very difficult exercises).

I'm having trouble grasping the following theorem:

Let S1 and S2 be collections of subsets of a set X such that X = (the union of all members of S1) = (the union of all members of S2). Moreover, suppose that:

1.) for each s1 in S1 and each x in s1, there exists an s2 in S2 such that x is in s2 and s2 is a subset of s1

2.) for each s2 in S2 and each x in s2, there exists an s1 in S1 such that x is in s1 and s1 is a subset of s2

Then S1 and S2 are subbases for the same topology on X.

My issue is that, this definition seems to imply that S1 and S2 are the exact same collection, I can't come up with two subcollections of X that satisfy this theorem and aren't the same.

For example, take X = {1,2,3,4,5} and

S1 = {{1},{1,2,3},{2,3,4},{3,5}}

I'm trying to construct an S2 that would generate the same topology on X, but I can't figure out how to come up with one that isn't exactly the same as S1.

If I consider {1}, then by the theorem I know that there must be a set in S2 that contains 1 and is a subset of {1}, which has to be {1}. So:

S2 = {{1}.....

Next, I consider {1,2,3}. I know there must be a set in S2 that contains 1 and is a subset of {1,2,3}. There already is, so I consider 2. There must be a set in S2 that contains 2 and is a subset of {1,2,3}. But I also know that this member of S2 must also be a subset of a set that contains 1, 2, and 3 in S1, but then they have to be the same set.

S2 = {{1},{1,2,3}... etc.

In ##\mathbb{R}^2##, take ##S_1## the set of open balls, that is:

$$\{(x,y)\in \mathbb{R}^2~\vert~\sqrt{ (x-a)^2 + (y-b)^2 }< r\}$$

Take ##S_2## the set of all open cubes, that is:

$$(a,b)\times (c,d)$$

These collections are not equal, but they induce the same topology.

I still don't see how the theorem allows them to differ, if they are subbases that generate the same topology.

WannabeNewton
There is nothing that says two subbases that generate the same topology must necessarily be equal; two bases that generate the same topology don't even have to be equal. Micromass gave the classic example of the ##\infty##-norm induced basis and the ##2##-norm induced basis; they both generate the usual Euclidean topology but they are not equal as sets. In fact all norms on a finite dimensional real/complex vector space, along with the associated bases of open balls, generate the same topology on that vector space.

I understand that they don't have to be equal (indeed that's why I know I don't' understand the theorem correctly in the first place), it's just that I can't interpret the theorem I posted having any other meaning. It essentially says that if I pick some element in subbasis A, then there is an element in B such that they are subsets of each other. So they are equal. What about the theorem am I not understanding?

Or in my example, how could I manufacture a subbasis that generates the same topology as the one I already have without making it the same?

I understand that they don't have to be equal (indeed that's why I know I don't' understand the theorem correctly in the first place), it's just that I can't interpret the theorem I posted having any other meaning. It essentially says that if I pick some element in subbasis A, then there is an element in B such that they are subsets of each other. So they are equal. What about the theorem am I not understanding?
No, not subsets of each other.

You know that for any ##A\in \mathcal{A}##, there is a ##B\in \mathcal{B}## such that ##B\subseteq A##.
And for any ##B\in \mathcal{A}##, there is an ##A^\prime\in \mathcal{A}## such that ##A^\prime\subseteq B##.

But nobody says that ##A = A^\prime##!!

I don't really have a problem with that, I think the part that is mucking it up for me is the "for each x in A."

So if I have ##A##={1,2,3} then not only is there a ##B## such that ##B## is a subset of ##A##, but there are ##B## that is a subset of ##A## that also contain "each x in A."

If I say "ok, these ##B## are {1}, {2}, {3}", then fine, the first part of the theorem holds because for every x in ##A##, there is a ##B## that is a subset of ##A## that contains x, but it doesn't work the other way around.

For it to work the other way around, it would have to be true that for the ##B## {1}, there is an ##A## that contains 1 that is a subset of {1}, which is {1}. So ##\mathcal{A}## must have to contain {1}. Since this same reasoning holds for {2} and {3}, ##\mathcal{A}## has to contain {1}, {2}, and {3}. So, that doesn't make ##\mathcal{A}## and ##\mathcal{B}## necessarily equal, but it does necessarily make ##\mathcal{B}## a subset of ##\mathcal{A}##.

The problem is that you're working with finite topological spaces. Things are going to make more sense if you consider infinite topological spaces such as ##\mathbb{R}^2##. Try to work with my suggestion in post 84. This is going to make more sense to you than finite topological spaces.

The problem is that you're working with finite topological spaces. Things are going to make more sense if you consider infinite topological spaces such as ##\mathbb{R}^2##. Try to work with my suggestion in post 84. This is going to make more sense to you than finite topological spaces.

Oh, okay. I'll try doing that. It usually helps me a lot to look a little simple examples I construct, to see why the theorems are true and "witness them" in that way, so it's a bit challenging when all of these simple examples I can do seem rather "degenerate."

But, just so we're super clear, when you say infinite topological spaces, what exactly does that mean? (is the set the topology is on of infinite cardinality, or is the topology of infinite cardinality, or are members of the topology of infinite cardinality?)

WannabeNewton
It means that the set should be countably infinite or, even better, uncountable. Topologies on finite sets are not rich enough for you to get anything fruitful out of, as far as examples go; this is why you keep running into confusions.

Oh, okay. I'll try doing that. It usually helps me a lot to look a little simple examples I construct, to see why the theorems are true and "witness them" in that way, so it's a bit challenging when all of these simple examples I can do seem rather "degenerate."

But, just so we're super clear, when you say infinite topological spaces, what exactly does that mean? (is the set the topology is on of infinite cardinality, or is the topology of infinite cardinality, or are members of the topology of infinite cardinality?)
Usually it means that the space ##X## is infinite, not that the topology is infinite. However, in this case, if ##\mathcal{T}## is finite, then the examples are still degenerate.

I know you don't like it, but topology really is most important when things are infinite. Unlike group theory or linear algebra, where we can see important concepts in the finite (finite-dimensional) case. Topology is very different.

It's not that I don't like it, I just need to adapt. :)

Thanks again guys

For example, you have no idea even what a generic open set in ##\mathbb{R}^2## looks like!
At least intuitively, an arbitrary open set in ##\mathbb{R}^2## seems like it would just be a countable union of regions bounded by piecewise contnuous functions. Is it more complicated than that?

At least intuitively, an arbitrary open set in ##\mathbb{R}^2## seems like it would just be a countable union of regions bounded by piecewise contnuous functions. Is it more complicated than that?
OK, but continuous curves in the plane can be very very wild. The difficulty in proving the Jordan curve theorem shows this.

But what I meant is open balls ad rectangles are very well-behaved objects. We know exactly how they look like. We certainly do have an intuition how open sets look like, but I don't think we can ever give an explicit description.

OK, but continuous curves in the plane can be very very wild. The difficulty in proving the Jordan curve theorem shows this.
Point taken; for instance space-filing curves. But is my characterization of open sets correct?

Point taken; for instance space-filing curves. But is my characterization of open sets correct?
I don't see any obvious counterexamples. But I can't really produce a proof at this moment.

I don't see any obvious counterexamples. But I can't really produce a proof at this moment.
I just started a thread about it here, where I try to phrase the question more formally.