Some introductory Topology questions

[1MileCrash]

Yes, I looked back and had the discrete and indiscreet topologies mixed up.

Pretty weird that they don't explain what it means for a topology to be generated by a metic.

Anyway, given a metric space $(X,d)$, we can define the open ball centered in $x\in X$ with radius $r>0$ as

B(x,r) := \{y\in X~\vert~d(x,y) < r\}

Now, a subset $G\subseteq X$ is defined to be open in the metric space if for any pont $x\in G$, there is an $r>0$ such that $B(x,r)\subseteq G$. I assume these are all concepts you've already seen??

Yes, my book has defined open balls. From what I can tell, it is just a set of y such that the "distance" between x and y is less than r.

The definition of open sets you gave was also given, but it's clear now that I should have spent more time on them, I'm having trouble getting a grasp over it.

It says that G is open if for any x in G, there is an r > 0 such that B(x,r) is a subset of G. So if for any x, there is an r such that the set of y such that the "distance" between x and y is less than r is a subset of G.

That is really confusing to me, I can't really make sense of it. I already understand open sets as a layman, in that it just "doesn't contain its boundary points" but I can't comprehend it in terms of open balls very well.

So, I can pick any element x of G,
and can find an r such that a set of y such that d(x,y) < r
is a subset of G.

Never before have I been able to type something and not "see it" or understand really what it implies at such a high degree. I really have no idea what this says about G.

Anyway, take your metric space $(X,d)$. The topology generated by the metric space is simply the collection of all open sets.

Ok

How would one do this?

Consider all subsets of X and then see it were true that "I can pick any element x of G, and can find an r such that a set of y such that d(x,y) < r is a subset of G." for each one?

Like, for the set X={1,2,3}
{{1}, {2}, {1,2},{2,3},∅, X}}

is a topology. So each element is an "open set."

Can you show me what it means to say that:

"I can pick any element x of G, and can find an r such that a set of y such that d(x,y) < r is a subset of G."

for, say, {1}?

I just don't get it.

 

[WannabeNewton]

Can you visualize how open sets in $\mathbb{R}^{2}$ are all unions of open balls under the Euclidean topology? Can you picture an open ball in $\mathbb{R}^{2}$ under such a topology?

Don't feel bad about not being able to visualize topology in full generality! It's by no means easy and often times impossible. I find it usually helps to visualize things in euclidean 2-space. It will all come in due time :)

 

[micromass]

Yes, I looked back and had the discrete and indiscreet topologies mixed up.



Yes, my book has defined open balls. From what I can tell, it is just a set of y such that the "distance" between x and y is less than r.

The definition of open sets you gave was also given, but it's clear now that I should have spent more time on them, I'm having trouble getting a grasp over it.

It says that G is open if for any x in G, there is an r > 0 such that B(x,r) is a subset of G. So if for any x, there is an r such that the set of y such that the "distance" between x and y is less than r is a subset of G.

That is really confusing to me, I can't really make sense of it. I already understand open sets as a layman, in that it just "doesn't contain its boundary points" but I can't comprehend it in terms of open balls very well.

So, I can pick any element x of G,
and can find an r such that a set of y such that d(x,y) < r
is a subset of G.

Never before have I been able to type something and not "see it" or understand really what it implies at such a high degree. I really have no idea what this says about G.

The problem with the boundary interpretation is that you have not yet defined what a boundary is. You can define it precisely in function of open sets. So you can define the boundary using open sets, and then the boundary of an open set will not intersect the open set. But I guess this is not helpful, since it requires you to define open sets first.

My advice would be to find some examples of open sets in $\mathbb{R}$ and $\mathbb{R}^2$. And to verify explicitely that they're open. This would already help a lot.

The open ball interpretation says that if you take a point $x$ in the open set, then there is an entire collection of points near $x$ that is also in the open set. So in any "direction" you go from $x$, you will always encounter points in the open set and not other points (unless you go too far). This makes "not including the boundary" rigorous. Because if you were on the boundary, then you could go into a certain direction and not encounter any points in the set. Openness guarantees that there is no such point in the set. So no points in the set are in the boundary.

How would one do this?

Consider all subsets of X and then see it were true that "I can pick any element x of G, and can find an r such that a set of y such that d(x,y) < r is a subset of G." for each one?

Like, for the set X={1,2,3}
{{1}, {2}, {1,2},{2,3},∅, X}}

is a topology. So each element is an "open set."

Can you show me what it means to say that:

"I can pick any element x of G, and can find an r such that a set of y such that d(x,y) < r is a subset of G."

for, say, {1}?

I just don't get it.

The topology you give does not come from a metric space. So I cannot interpret it as function of open balls and metrics.

 

[1MileCrash]

Thanks for the advice you two, I am going to sleep on it and try some more of your suggestions and do some thinking on it tomorrow.

 

[WannabeNewton]

Sleep on it and compile a list of all the questions you have left! Have a good night.

 

[1MileCrash]

*sigh* I think I understand the ball definition of open sets, finally. All I had to actually draw a circle, call it a set, look at the definition, and I realize I was making it much more complicated than it needed to be.

Basically, it says that I can pick any element x in the set G, and I know that there are other elements in the set G that differ from x by a nonzero amount r. Even though the difference r is "radially," the important part is that it includes those elements that are nearer to the boundary of the set. So it's just saying that for any element x, there are elements that are closer to the boundary by a nonzero amount.

Is that right?

Even if I do understand that bit, it is still hard to imagine this making sense with finite sets.

 

[WannabeNewton]

What it's saying is that if you take any $x \in G$, where $G$ is an open set in a metric space, we can always find a sufficiently small radius $r$ such that all points within a distance $r$ from $x$ lie within $G$. This is exactly what you said if we take for granted that the notion of "boundary" has been defined. It's not as "geometric" with arbitrary topologies on finite sets, I agree, but picturing things in metric spaces can still help you a lot. By the way, all finite metric spaces are discrete so that should help you picture them better (try it).

 

[1MileCrash]

Ok, I'm still having a really hard time linking up the metric space and topological space.

They sound like two completely different ideas.

My book says that "every metric space is a topological space" but how does that make sense? A metric space is a set and a metric, which is a distance function. A topological space is a set and a collection of open subsets of that set.

If a metric space is also a topological space, then that would mean that this function d is a collection of subsets. How can you call a function with a numerical output a collection of subsets? How are these things equivalent?

 

[Fredrik]

My book says that "every metric space is a topological space" but how does that make sense?

Strictly speaking, that statement isn't true. But given a metric space (X,d), you can always define a topology on X by saying that a subset of X is open if and only if it's a it's union of open balls in X, with respect to d. If we define $\tau$ as the set of all subsets of X that are "open" in this sense, then $(X,\tau)$ is a topological space.

This is what people mean when they say that every metric space is a topological space.

 

[WannabeNewton]

If a metric space is also a topological space, then that would mean that this function d is a collection of subsets. How can you call a function with a numerical output a collection of subsets? How are these things equivalent?

As Fredrik said, the open balls of the metric form a topology generating basis i.e. the topology is that which is generated by the basis of open balls of the metric.

 

[1MileCrash]

Ok, I came up with this example for myself to try to understand the process, and to make my questions more clear.

Let X = { 1, 2, 3, 4, 5}
Let d be the normal metric d(x,y) = |x-y|
So (X, d) is a metric space.

I want to generate the induced topology. I know that the topology will be the collection of all open subsets of X.

How do I find all open subsets of X?

I listed every subset of {1,2,3,4,5} but I don't understand how to check if any particular subset is open.

Take for example, {1,2}

I know that this set is open if

"Now, a subset $G\subseteq X$ is defined to be open in the metric space if for any pont $x\in G$, there is an $r>0$ such that $B(x,r)\subseteq G$."

There are two points in this subset to consider, 1 and 2.

B(1,r) has to be a subset of {1,2} for some r > 0.

B(1,r) = {y: |1-y| < r}

let r = .00000042

Then B(1,r) = {1}

and that's a subset of {1,2}

B(2,r) has to be a subset of {1,2} for some r>0

B(2,r) = {y: |2-y| < r}

let r = .000000042

Then B(1,r) = {2}

and that's a subset of {1,2}

So {1,2} is an open set.

What did I just say? It seems like I could make any set that I wanted open. Why can't I just always make r small enough so that B is always just the singleton set of x itself, which is of course a subset of the set I'm considering, because I picked it from the set? It's just not making any sense to me at all.

Now, a subset $G\subseteq X$ is defined to be open in the metric space if for any pont $x\in G$, there is an $r>0$ such that $B(x,r)\subseteq G$. I assume these are all concepts you've already seen??

Anyway, take your metric space $(X,d)$. The topology generated by the metric space is simply the collection of all open sets.

This helps me understand what a topology is, and what an open set is, but I still don't see how I generate a topology from a metric. My only guess is that I have to list every single subset of X, and then look and see if there is an r such that the ball of radius r is a subset of X for every single element of every single subset of X, which according to my great lack of understanding above, there will be an r no matter what that satisfies that idea. This seems completely impossible to do anyway, my first exercise is to prove that the discrete metric generates the discrete topology, but I'm not told anything about the set its on, so what I've just described can't be the process. What is the process of generation?

 

[micromass]

Ok, I came up with this example for myself to try to understand the process, and to make my questions more clear.

Let X = { 1, 2, 3, 4, 5}
Let d be the normal metric d(x,y) = |x-y|
So (X, d) is a metric space.

I want to generate the induced topology. I know that the topology will be the collection of all open subsets of X.

How do I find all open subsets of X?

I listed every subset of {1,2,3,4,5} but I don't understand how to check if any particular subset is open.

Take for example, {1,2}

I know that this set is open if

"Now, a subset $G\subseteq X$ is defined to be open in the metric space if for any pont $x\in G$, there is an $r>0$ such that $B(x,r)\subseteq G$."

There are two points in this subset to consider, 1 and 2.

B(1,r) has to be a subset of {1,2} for some r > 0.

B(1,r) = {y: |1-y| < r}

let r = .00000042

Then B(1,r) = {1}

and that's a subset of {1,2}

B(2,r) has to be a subset of {1,2} for some r>0

B(2,r) = {y: |2-y| < r}

let r = .000000042

Then B(1,r) = {2}

and that's a subset of {1,2}

So {1,2} is an open set.

What did I just say? It seems like I could make any set that I wanted open. Why can't I just always make r small enough so that B is always just the singleton set of x itself, which is of course a subset of the set I'm considering, because I picked it from the set? It's just not making any sense to me at all.

Right, any set is open here. So you could do the same with any set. Like already has been said, all finite metric spaces have the property that all sets are open. So if you want nontrivial examples, you need to look at infinite spaces.

This helps me understand what a topology is, and what an open set is, but I still don't see how I generate a topology from a metric.

Usually, you don't find the explicit topology at all, it's just too complicated. For example, you have no idea even what a generic open set in $\mathbb{R}^2$ looks like! However, you do know what the open balls look like. And the open balls generate the topology (in the sense that any open set is a union of open balls). So we say that the open balls form a basis. A lot of topological concept (actually, almost all) can be stated for bases instead of topologies. So the important concept in topology is actually a basis. A topology is usually too big.

My only guess is that I have to list every single subset of X, and then look and see if there is an r such that the ball of radius r is a subset of X for every single element of every single subset of X, which according to my great lack of understanding above, there will be an r no matter what that satisfies that idea. This seems completely impossible to do anyway, my first exercise is to prove that the discrete metric generates the discrete topology, but I'm not told anything about the set its on, so what I've just described can't be the process. What is the process of generation?

The set is arbitrary. So take an arbitrary set $X$ with the discrete metric. Now prove that any set is open.

Again, if you want more nontrivial examples, then look at open sets in $\mathbb{R}^2$.

 

[1MileCrash]

Thanks for your patience mm, so you mean that any subset of any finite set is always open regarding any metric on that set?

And, does this mean I am guaranteed to have a topology on a finite set if and only if I:

1.) Put the empty set and the entire finite set itself in the collection
2.) Take however many of whatever subsets I want, and put those in (the "basis?")
3.) Put the union/intersection of any number of any of the subsets I put in during (2), and put those in as well

But that's not true of infinite metric spaces because I couldn't just take whichever subset I wanted during (2), I'd have to verify it was open first, and then I could proceed in the same way?The next section is about topological basis. Do you think reading on would help me, or should I cover this better first?

The set is arbitrary. So take an arbitrary set $X$ with the discrete metric. Now prove that any set is open.

So it would be like,

Let (X,d) be the discrete metric space.

Let G be a subset of X, and g be an element of G.

G is open if there exists an r such that

B(g,r) = {y : d(g,y) < r } is a subset of GSo, I've tried proceeding here and I don't know what to do. Normally when I'm showing that something is a subset of something else, I pick an arbitrary element of the set and show it's in the other one.

Here I have to both show that any element of {y : d(g,y) < r } is a subset of G and show that there is an r that can make that happen. It's like I need one to prove the other, I don't see how to do these two things at once.

Is it the case that since d(g,y) is always 0 or 1, that I just need to show that any ball of radius > 1 is a subset of G? But wouldn't that mean that G would have to contain more than one element, thus not making G arbitrary?

Ugh, I'm so confused.

 

[micromass]

Thanks for your patience mm, so you mean that any subset of any finite set is always open regarding any metric on that set?

Yes.

And, does this mean I am guaranteed to have a topology on a finite set if and only if I:

1.) Put the empty set and the entire finite set itself in the collection
2.) Take however many of whatever subsets I want, and put those in (the "basis?")
3.) Put the union/intersection of any number of any of the subsets I put in during (2), and put those in as well

For metric spaces, you know that every set is open. So you need to end up with a topology that is every set. I don't see how your process guarantees that everything is open.

The next section is about topological basis. Do you think reading on would help me, or should I cover this better first?

I think you should verify some easy examples of topologies. Are there any exercises after this section? If not, I can provide you with some good exercises.

So it would be like,

Let (X,d) be the discrete metric space.

Let G be a subset of X, and g be an element of G.

G is open if there exists an r such that

B(g,r) = {y : d(g,y) < r } is a subset of G


So, I've tried proceeding here and I don't know what to do. Normally when I'm showing that something is a subset of something else, I pick an arbitrary element of the set and show it's in the other one.

Here I have to both show that any element of {y : d(g,y) < r } is a subset of G and show that there is an r that can make that happen. It's like I need one to prove the other, I don't see how to do these two things at once.

Is it the case that since d(g,y) is always 0 or 1, that I just need to show that any ball of radius > 1 is a subset of G? But wouldn't that mean that G would have to contain more than one element, thus not making G arbitrary?

So, you take $G$ arbitrary. You take $g\in G$ arbitrary. You need to find an $r$ such that $\{y~\vert~d(g,y)<r\}\subseteq G$. Focus on finding the right $r$ first.

Now, as a hint, for a given $r$, what are all the possibilities for $\{y~\vert~d(g,y)<r\}$. You will need to look at the cases $r\leq 1$ and $r>1$.

 

[1MileCrash]

Yes.



For metric spaces, you know that every set is open. So you need to end up with a topology that is every set. I don't see how your process guarantees that everything is open.

Sorry, I was a bit confused. I thought we had said that any subset of a finite metric space was automatically open, I need to reread.

I think you should verify some easy examples of topologies. Are there any exercises after this section? If not, I can provide you with some good exercises.

There are, I just don't think I'm in the position to attempt them. The last section, they started off really simple and got difficult, but I could do them all with time. The first one here, is a brick wall.

So, you take $G$ arbitrary. You take $g\in G$ arbitrary. You need to find an $r$ such that $\{y~\vert~d(g,y)<r\}\subseteq G$. Focus on finding the right $r$ first.

Now, as a hint, for a given $r$, what are all the possibilities for $\{y~\vert~d(g,y)<r\}$. You will need to look at the cases $r\leq 1$ and $r>1$.

Let r <= 1, then
d(g,y) < 1.

Since d is the discrete metric, that means that g = y and thus B(g,r) = {g}, which we know to be a subset of G since g is an element of G.

The other case is a bit confusing, I don't think what I say really makes sense.

--Ignore--

Let r > 1, then we have that

B(g,r) = {y : d(g,y) < r}

But if r is larger than 1, then the entire set of X is a subset of it, because the maximum distance allowed by the discrete metric is 1. Since X is a subset of B(g,r) if r > 1, then so is G.

WAIT, I need to show that B(g,r) is a subset of G, not the other way around. >< Ignore for now!

___

Can I say that the ball with r > 1 is "bigger" than X itself?

 

[micromass]

Sorry, I was a bit confused. I thought we had said that any subset of a finite metric space was automatically open, I need to reread.

Yes, any subspace of a finite metric space is open. So if $(X,d)$ is a finite metric space, then $\mathcal{T}= \mathcal{P}(X)$.

Let r <= 1, then
d(g,y) < 1.

Since d is the discrete metric, that means that g = y and thus B(g,r) = {g}, which we know to be a subset of G since g is an element of G.

Good. Doesn't this solve the exercise already?

The other case is a bit confusing, I don't think what I say really makes sense. If the radius of the ball is bigger than any d, can I say that the "boundary" of the ball lies outside of the set that the ball is in, so the set doesn't contain its boundary, so it is open?

Sorry, I have no idea what you're trying to say here.

Let r > 1, then we have that

B(g,r) = {y : d(g,y) < r}

But if r is larger than 1, then the entire set of X is a subset of it, because the maximum distance allowed by the discrete metric is 1. Since X is a subset of B(g,r) if r > 1, then so is G.

OK, so $X=B(g,r)$ in this case.

 

[1MileCrash]

Good. Doesn't this solve the exercise already?

Yeah, it does, because d is always <= 1.

So any ball for this metric is a singleton set for the center of the ball?

Sorry, I have no idea what you're trying to say here.

You can ignore it, I'm still forming thoughts and thinking out loud, it's a mess up there right now.

OK, so $X=B(g,r)$ in this case.

So clearly, we can't say that B(g,r) is a subset of G. But, then again, there is no such g,y such that d(g,y) > 1, so does it matter?

 

[micromass]

Yeah, it does, because d is always <= 1.

What do you mean? $d$ is a function, how can it be smaller than 1?

So any ball for this metric is a singleton set for the center of the ball?

Any ball with radius smaller or equal than $1$ is. If the radius is strictly larger than $1$, than the ball is the entire space.

So clearly, we can't say that B(g,r) is a subset of G. But, then again, there is no such g,r such that d(g,r) > 1, so does it matter?

What do you mean with $d(g,r)$?? I thought $r$ was a number. In that case $d(g,r)$ makes no sense.

 

[1MileCrash]

What do you mean? $d$ is a function, how can it be smaller than 1?

I mean that d(x,y) is less than or equal to 1 for any x,y. So why do we need to consider balls with a radius larger than the distance between any two points?

What do you mean with $d(g,r)$?? I thought $r$ was a number. In that case $d(g,r)$ makes no sense.

Sorry, that should be d(g,y).EDIT:

Darn, I just realized what you're saying. I need to affirm that an r EXISTS, and by showing that any r < 1 works for any arbitrary x in G, I've done just that. That's why the exercise is done.

 

[micromass]

I mean that d(x,y) is less than or equal to 1 for any x,y. So why do we need to consider balls with a radius larger than the distance between any two points?

We don't need to consider them for this problem. But it's still interesting to know what the balls are. So something like $B(g,2)$ exists here (and this set is equal to $X$). But it's not interesting. As you will soon find out, in topoogy we only care about balls with "small" $r$.

 

[micromass]

Darn, I just realized what you're saying. I need to affirm that an r EXISTS, and by showing that any r < 1 works for any arbitrary x in G, I've done just that. That's why the exercise is done.

OK, you got it, I think.

 

[1MileCrash]

We don't need to consider them for this problem. But it's still interesting to know what the balls are. So something like $B(g,2)$ exists here (and this set is equal to $X$). But it's not interesting. As you will soon find out, in topoogy we only care about balls with "small" $r$.

That's clear, thanks.

 

[1MileCrash]

I'm wondering if my book's definition for "subbasis" makes sense to any of you? It doesn't seem to tell me anything about a subbasis besides it being a subcollection of a topology with the mentioned property (about the topology, not the subcollection). Is it a typo maybe?

 

[WannabeNewton]

Yes it seems to be a typo. Here's the definition from Lee: http://postimg.org/image/91fd5oc8t/

 

[1MileCrash]

Wow, this book is plagued with errors. So, is the typo that the second to last T should actually be an l (the sub collection)?

And another question about the incorrect definition. Can a topology whose finite intersections form a basis for itself exist? It seems like finite intersections of members of a topology would always end up excluding elements that exist in members of the topology, and thus reforming the topology with unions (as in a basis) couldn't happen. If there's at least one member of the topology with an element that no other member has, this isn't possible. Right?

Just something out of curiosity.

 

[jgens]

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.

 

[1MileCrash]

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?

 

[1MileCrash]

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.

 

[1MileCrash]

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.

 

[micromass]

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).

 

[1MileCrash]

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.

 

[micromass]

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 }&lt; 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.

 

[1MileCrash]

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.

 

[1MileCrash]

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?

 

[micromass]

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 $A = A^\prime$!

 

[1MileCrash]

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 $\mathcal{A}$.

 

[micromass]

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.

 

[1MileCrash]

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.

 

[micromass]

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.

 

[1MileCrash]

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

Thanks again guys

 

[lugita15]

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?

 

[micromass]

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.

 

[lugita15]

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?

 

[micromass]

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.

 

[lugita15]

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.