What is the Significance of Completely Regular Spaces?

[friend]

Yes, but you don't choose a particular y in F. The continuous function must be 1 for all y in F.

It also seems in the definition you have "for any point x not in F", f(x)=0. So it seems that no matter what point x in the topology not in F, f(x)=0. In other words, the function f must always be zero outside F and 1 inside F, since you can chose x anywhere outside F, and yet must have f(x)=0? Or are we talking about a different continuous function f for each choice of x?

Then if f(x)=0 for all points x outside F, there seems to be a discontinuity where if you approach the edge of F from outside f=0, but if you approach the edge from inside F, then f=1. This sounds like the definition of discontinuous.

 

[dx]

We are talking about different functions for each choice of x and F.

Once you choose x and F, then we must be able to find a continuous function that separates them. This function only needs to be 0 at x, not anywhere else.

 

[WannabeNewton]

Then if f(x)=0 for all points x outside F, there seems to be a discontinuity where if you approach the edge of F from outside f=0, but if you approach the edge from inside F, then f=1. This sounds like the definition of discontinuous.

Your intuitive notions of continuity won't help here so don't use it. Even in \mathbb{R} there are many textbook maps that are continuous in a counter intuitive manner. For example: http://en.wikipedia.org/wiki/Popcorn_function

Try to show this is continuous in the manner specified in the wiki article using the epsilon delta definition of continuity, it will be very instructive in allowing you to get a hold of continuity (but it isn't an easy problem so it might take some time!). In topology, continuity is not as trivial in an intuitive sense as it is for the nice functions one sees in physics.

 

[friend]

We are talking about different functions for each choice of x and F.

Once you choose x and F, then we must be able to find a continuous function that separates them. This function only needs to be 0 at x, not anywhere else.

Even if x approaches the boundary of F?

 

[dx]

Why would that be a problem? x can never be on the boundary of F since closed sets contain their boundaries, but x can be any point outside F.

 

[Bacle2]

Friend, following up upon dx's last post: have you tried to see if the result holds for F open?

Also notice, like dx said, for every choice of F and x there is a function with the given
properties; not for a given F closed and _all x_, but, given F and given a specific x not
in F . If you have a choice of F and x, you can find a function, say, f(F,x) that will have
the given properties; if you choose a different pair F', x' , you will (except maybe in very exceptional situations) a different f'(F'x').

 

[friend]

Why would that be a problem? x can never be on the boundary of F since closed sets contain their boundaries, but x can be any point outside F.

So does continuity say in other words that no matter how close you get to the boundary of F, the open nature of the entire set minus the closed set F will always allow you to find an open set even closer to the boundary that maps to an open interval even closer to 1? Though, I suppose the strict definition would require the inverse map, from the open intervals to the open sets in the topology.

 

[dx]

You can say it this way:

Since f is continuous, and since f is 1 on F, f can be made as close to 1 as you want if you evaluate it at points sufficiently close to F

In fact, in terms of neighborhoods, a map f is continuous if and only if given any point x and any neighborhood M of f(x), there is a neighborhood N of x such that f[N] is a subset of M.

 

[friend]

You can say it this way:

Since f is continuous, and since f is 1 on F, f can be made as close to 1 as you want if you evaluate it at points sufficiently close to F

In fact, in terms of neighborhoods, a map f is continuous if and only if given any point x and any neighborhood M of f(x), there is a neighborhood N of x such that f[N] is a subset of M.

Very good! Thank you. Let's move on...

A set is a closed if its complement is open, so the open sets determine what the closed sets are. And its not unique. We are not talking about some particular closed set F. For any closed set F, and any point x outside it, there must be a continuous function that separates them.

We've talked about what it means to say "any point x not in F". Now I would like to go back and understand "any closed set F". I understand that there is a whole space of points from which to arbitrarily choose any x, well, except perhaps x can not be chosen from F. But I'm not seeing how many choices we have from which to create any F.

I'm trying to avoid any appearance of uniqueness in F so that the property of complete regularity encompasses the whole topology. Otherwise, with x restricted from being chosen in F, it seems there are restricted regions for which the definition does not apply, which means the property is not applicable to the whole topology. So it seems necessary that x can be chosen from any point in the topology and we are guaranteed to be able to find an F and continuous functions f to comply with the definition. So I don't think the definition means draw a circle anywhere you like and call it F. Your comment suggests that F must be constructed from the complement of the available open sets in the topology. So it seem the topology allows more than one F so we can choose x in the previous F and still find a different F that fulfills the definition. Does this sound right?

 

[dx]

Yes, a set is closed only if its complement is an open set, i.e. a member of the topology. Given a certain closed set F, then points outside F must be separated from F by a continuous function. The points here are outside F, but they don't have to be outside other closed sets. For a different closed set F', there must be continuous functions that separate F' from points outside F', and again these points are all points outside F', whether they belong to closed sets other than F' or not.

 

[friend]

Yes, a set is closed only if its complement is an open set, i.e. a member of the topology. Given a certain closed set F, then points outside F must be separated from F by a continuous function. The points here are outside F, but they don't have to be outside other closed sets. For a different closed set F', there must be continuous functions that separate F' from points outside F', and again these points are all points outside F', whether they belong to closed sets other than F' or not.

Thank you. I'm finding it easier to understand your explanations. So is there any x anywhere in the entire topology for which we may not be able to find an F that fulfills the definition?

Does the definition also apply to subsets of the topology (subspaces?). Say we divide the topology in half. Would we be able to find x, F, and f in one half that fulfill the definition?

 

[dx]

Thank you. I'm finding it easier to understand your explanations. So is there any x anywhere in the entire topology for which we may not be able to find an F that fulfills the definition?

The definition does not say anywhere that you must be able to find an F and an x. It only says that if F is a closed set, and x is a point outside F, then there is a continuous function that separates them.

Does the definition also apply to subsets of the topology (subspaces?). Say we divide the topology in half. Would we be able to find x, F, and f in one half that fulfill the definition?

If a topological space is completely regular, then subspaces of that space are also completely regular. You could try proving that.

 

[friend]

The definition does not say anywhere that you must be able to find an F and an x. It only says that if F is a closed set, and x is a point outside F, then there is a continuous function that separates them.

If a topological space is completely regular, then subspaces of that space are also completely regular. You could try proving that.

I suspect that if a completely regular topological space can be separated into subspaces that are also completely regular, then it must have been true that in the whole topological space for any x whatsoever there must be an F and an f that fulfills the definition. For then the whole topological space could be divided into subspaces where each has a different F that allowed all x in the definition of CR in the original space.

But if you could point me to the proof of where these subspaces carry the same property, that might be worth buying one of those $100 books. Thank you.

 

[micromass]

Can we please stay on topic?

I don't like friend his behaviour either. And if it were up to me, I would have locked the thread. But dx specifically asked me to open the thread again so he could help further. I don't want this thread to turn into another shouting match.

If anybody wishes to further aid the OP, then please do so. Otherwise, I would ask to stay out of this thread.

Any off-topic posts will be deleted. Comments on this post are welcome through sending a PM or through a report.

Snarky comments will be warned/infracted. In particular, things like "Are you just guessing?" are not allowed.

 

[jbunniii]

But if you could point me to the proof of where these subspaces carry the same property, that might be worth buying one of those $100 books. Thank you.

You don't need to spend $100. Willard's General Topology covers all of this and has a retail price of US$22.95 (currently only $14.70 from Amazon). As a bonus, it contains the nice quote: "A counterexample exists showing that not every regular space is completely regular. It is formidable and we have relegated it to Exercise 18G, where most people won't be bothered by it."

 

[friend]

You don't need to spend $100. Willard's General Topology covers all of this and has a retail price of US$22.95 (currently only $14.70 from Amazon). As a bonus, it contains the nice quote: "A counterexample exists showing that not every regular space is completely regular. It is formidable and we have relegated it to Exercise 18G, where most people won't be bothered by it."

Yea, Dover can publish some pretty good stuff. I looked at it on Amazon.com and order it. Maybe after reading that book, I'll be able to contribute more constructive here instead of just asking a lot of questions. Thank you.

 

[WannabeNewton]

I have to disagree with my significant other, jbunnii, on this one. Yes the price is cheap but Willard is an extremely tough book. Proceed with caution is all I can say.

 

[friend]

I have to disagree with my significant other, jbunnii, on this one. Yes the price is cheap but Willard is an extremely tough book. Proceed with caution is all I can say.

Thanks for the warning. Maybe my previous experience with topology will prevent any shocks along the way. I feel like I'm sufficiently motivated to take up the subject. But I hope you all won't mind if I have to ask some questions along the way.

I remember reading "Tensor Analysis on Manifolds" by Bishop and Goldberg by Dover. That was pretty intense too, but I was able to follow it and appreciated the rigor. Of course, there was certainly a lot of highlighting and underlining and had to read it more than once. If the topology book is like that, I think I will appreciate it.

 

[WannabeNewton]

There's nothing wrong with a challenge and it's cheap so you don't have anythin to lose really. I've never read Bishop so I can't say I can relate, any and all tensor calculus I know I learned secondhand from GR textbooks. Anyways, good luck with whatever you choose to do mate.

 

[jbunniii]

I have to disagree with my significant other, jbunnii, on this one. Yes the price is cheap but Willard is an extremely tough book. Proceed with caution is all I can say.

My better half WannabeNewton is right, Willard is by no means easy. I can recommend another nice Dover book, Gemignani's Elementary Topology, which is not as comprehensive as Willard but it is easier to read and the exercises are not nearly as demanding. It does cover complete regularity near the end. It's only $8.35 from Amazon, so you can't really go wrong. Willard is more fun to read, though.

 

[friend]

My better half WannabeNewton is right, Willard is by no means easy. I can recommend another nice Dover book, Gemignani's Elementary Topology, which is not as comprehensive as Willard but it is easier to read and the exercises are not nearly as demanding. It does cover complete regularity near the end. It's only $8.35 from Amazon, so you can't really go wrong. Willard is more fun to read, though.

From what I gather, this book introduces metric spaces kind of early. I was kind of hoping to get as much as I could out of point sets before introducing the added structure of a metric with its associated neighborhoods. The previous book seems to do that.

 

[WannabeNewton]

Metric spaces help motivate various ideas revolving around topological spaces. Many preliminary definitions in topology are motivated through similar theorems in metric spaces and usually allow for more elegant proofs, which one could appreciate if one has seen some stuff on metric spaces beforehand. For example the definition of a continuous map between two metric spaces is f:(X,d_1)\rightarrow (Y,d_2) is continuous if for all a\in X, for all \epsilon > 0, there exists a \delta >0 such that for all x\in X satisfying d_1(x,a) < \delta we have d_2(f(x),f(a)) < \epsilon. You can easily "picture" this intuitively as saying: give me an open ball of arbitrarily small radius \epsilon around f(a) and I can always find an open ball of radius \delta around a to match it.

When you see the definition of continuity in topology, where there is no notion of "smallness" due to the lack of a metric, the metric space definition provides great motivation and at least let's you see where the definitions come from because the similarities are clear. This build up of intuition is very helpful in a subject like topology.

 

[friend]

Metric spaces help motivate various ideas revolving around topological spaces. Many preliminary definitions in topology are motivated through similar theorems in metric spaces and usually allow for more elegant proofs, which one could appreciate if one has seen some stuff on metric spaces beforehand. For example the definition of a continuous map between two metric spaces is f:(X,d_1)\rightarrow (Y,d_2) is continuous if for all a\in X, for all \epsilon > 0, there exists a \delta >0 such that for all x\in X satisfying d_1(x,a) < \delta we have d_2(f(x),f(a)) < \epsilon. You can easily "picture" this intuitively as saying: give me an open ball of arbitrarily small radius \epsilon around f(a) and I can always find an open ball of radius \delta around a to match it.

When you see the definition of continuity in topology, where there is no notion of "smallness" due to the lack of a metric, the metric space definition provides great motivation and at least let's you see where the definitions come from because the similarities are clear. This build up of intuition is very helpful in a subject like topology.

You are of course welcome to your opinion. Personally, I feel there is no inherent necessity of a metric on point sets. So adding a metric only obscures the inherent nature of various properties being described. I'd rather avoid a metric until it is absolutely necessary.

 

[WannabeNewton]

It would only be beneficial is what I'm saying. It is sort of like killing two birds with one stone. I saw before that you had an interest in QM (correct me if I'm wrong); well functional analysis is the underlying mathematics of QM and functional analysis is heavy on metric spaces and topological vector spaces so, again, killing two birds with one stone. When you start seeing L^{2} in QM for example, these concepts will be important.

 

[friend]

It would only be beneficial is what I'm saying. It is sort of like killing two birds with one stone. I saw before that you had an interest in QM (correct me if I'm wrong); well functional analysis is the underlying mathematics of QM and functional analysis is heavy on metric spaces and topological vector spaces so, again, killing two birds with one stone. When you start seeing L^{2} in QM for example, these concepts will be important.

Yes, QM is the driving force behind my curiosity and many questions. Not to get off topic, but I noticed that some quantum mechanical effects could be derived from the Dirac delta function which seems to be an infinitismal version of the Dirac measure which is a function that returns a 1 if an element is included in a set and returns 0 if the element is not included in the set.

Then I noticed that completely regular spaces are defined with a very similar construction, a function that returns 1 for y elements of a closed set F. I had to investigate how this property is related to the whole space.

This makes me wonder if there are not other things in topology that return a numerical value depending on how points in a set are related. Would I be interested in the indicator function, for example? Is it used to define any properties of a topological space or a manifold?

But as far as metrics go, of course one ends up having to study them. But I'm wondering, is a metric something that is derived from topological properties? Or are they only imposed for other purposes?

 

[friend]

I do appreciate the help I'm receiving here on Physics Forums. And I have yet another question, hopefully an easy one.

In the definition of completely regular spaces, can a single point serve as the "closed set F"? IIRC, and I might be wrong, a single element set can be both open and closed. Is this right?

And the next question, can the function f be 0 for all points outside F and 1 only for points inside F? Or would that violate continuity? Thanks.

 

[dx]

In the definition of completely regular spaces, can a single point serve as the "closed set F"? IIRC, and I might be wrong, a single element set can be both open and closed. Is this right?

Yes, a single point can be closed. If the topological space is discrete, then all sets are open, and all sets are closed.

And the next question, can the function f be 0 for all points outside F and 1 only for points inside F? Or would that violate continuity? Thanks.

That is definitely possible. Again, consider a discrete space X. Then every map X → ℝ is continuous.

 

[friend]

Thank you very much, dx. I feel I'm nearing the end of my questions with your help.

Yes, a single point can be closed. If the topological space is discrete, then all sets are open, and all sets are closed.

Can a single point be a closed set in a non-discrete, connected space?

That is definitely possible. Again, consider a discrete space X. Then every map X → ℝ is continuous.

What about in a connected, non-discrete space?

 

[dx]

Can a single point be a closed set in a non-discrete, connected space?

Sure. Take ℝ, the space of real numbers, with the standard topology. This space is connected. Any point x in ℝ is a closed set.

What about in a connected, non-discrete space?

That is not possible. If f: X → ℝ is a continuous map, and X is connected, then f[X] must be connected in R. If f was 1 on some closed set, and 0 everywhere else, then f[X] = {0, 1}, which is not a connected subset of ℝ.

 

[friend]

Sure. Take ℝ, the space of real numbers, with the standard topology. This space is connected. Any point x in ℝ is a closed set.

That's great. I suppose my last question would be can every single-point closed set in the space serve in the definition of complete regularity? Or since the closed sets must be constructed from the complement of the available open sets in the topology, might the topology not allow some points to serve as the closed sets in the definition? If not every multi-point closed set can serve as the closed set in the definition, then I would suppose not every single-point set can serve in the definition.