What is the Significance of Completely Regular Spaces?

[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.

 

[dx]

If the complement of the set is open, then its a closed set. That's all there is to it.

The definition says that a particular condition must be satisfied for all closed sets in that space. If the topology says that some set is not a closed set, then you don't consider that set. The condition does not have to hold for such sets, since they are not closed.

 

[WannabeNewton]

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 type of construction is very common so don't wrack your head looking for deeper connections. As you mentioned, the indicator function is something that shows up everywhere simply because of its practical utility. For example in measure theory it is used in the definition of the lebesgue integral.

 

[friend]

This type of construction is very common so don't wrack your head looking for deeper connections. As you mentioned, the indicator function is something that shows up everywhere simply because of its practical utility. For example in measure theory it is used in the definition of the lebesgue integral.

Thanks WBN,

What I'm really interested in is those constructions that are necessary in the definition of manifolds and are allowed to be constructed everywhere on it. I would then look into whether those constructions might be useful in also constructing quantum mechanical effects everywhere on that manifold. I don't know if that can be done, but it certainly would be interesting to discover that manifolds necessarily carry QM structures, wouldn't it, perhaps in the form of virual particles?

 

[friend]

I've been thinking more about this. And I wonder if I have enough information to conclude that all manifolds automatically admit quantum mechanical structures. But I'd like your advice as to whether I'm making a mistake. Thank you.

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.

My reasoning is as follows:

1) "every topological manifold is Tychonoff", as stated here.

2) "X is a Tychonoff space... if it is both completely regular and Hausdorff", as stated here.

3) Therefore, every topological manifold is completely regular.

4) A completely regular space is defined by:
"X is a completely regular space if given any closed set F and any point x that does not belong to F, then there is a continuous function f from X to the real line R such that f(x) is 0 and, for every y in F, f(y) is 1. In other terms, this condition says that x and F can be separated by a continuous function." This is stated here again.

5) Since every point in a manifold is a "closed set", see quote from dx above, then F in the definition in 4) can be a single point. So every topological manifold admits a continuous function, f, from one point, x, to another point, y, such that f(x)=0, and f(y)=1.

6) The function described in 5) could be considered to be the integral of a probability distribution, which is 0 when integrated from one point to the same point, but is 1 when integrated from x to y, if that range encompasses all possibilities.

7) Since 5) and 6) must be true for all points x for a given y, even for x arbitrarily close to y, and must also be true for all points y since each point is a closed set required to be accommodated in the definition, then there must be a Dirac delta between any two points in the manifold, or at least a gaussian distribution between any two points even in the limit where x approaches y.

8) The Dirac delta, or even just a gaussian, can be manipulated into the Feynman Path Integral of quantum mechanics as I've posted many time in PF, for example, here. These manipulations work also for any gaussians because of the Chapman-Kolmogorov equation, which I can show if asked.

9) Therefore, every topological manifold is a completely regular space which must admit Dirac delta functions between any two points, which can be manipulated into the Path Integral of quantum mechanics. So every manifold necessarily includes a quantum mechanical structure in its definition.

Since this derivation has not been published anywhere as far as I know, I submit this derivation as a question to the members of PF, for your appraisal and criticism. I'm not stating this as a speculative theory; I'm here just to ask the PF members to help identify the issues in such a proof. Thank you.

 

[friend]

4) A completely regular space is defined by:
"X is a completely regular space if given any closed set F and any point x that does not belong to F, then there is a continuous function f from X to the real line R such that f(x) is 0 and, for every y in F, f(y) is 1. In other terms, this condition says that x and F can be separated by a continuous function." This is stated here again.

5) Since every point in a manifold is a "closed set", see quote from dx above, then F in the definition in 4) can be a single point. So every topological manifold admits a continuous function, f, from one point, x, to another point, y, such that f(x)=0, and f(y)=1.

Since a continuous function, f, must have a continuous domain, and since every point in the topology can be considered a closed set which must be accommodated in definition 4), it seems that a completely regular space is path connected from any point to any other point. Is that enough to specify a manifold? Or do you also need the property of being Hausdorff?

6) The function described in 5) could be considered to be the integral of a probability distribution, which is 0 when integrated from one point to the same point, but is 1 when integrated from x to y, if that range encompasses all possibilities.

Well actually, the function f in the definition in 4) above could go negative, as long as f(x)=0 and f(y)=1, right? Probability distributions don't go negative.

7) Since 5) and 6) must be true for all points x for a given y, even for x arbitrarily close to y, and must also be true for all points y since each point is a closed set required to be accommodated in the definition, then there must be a Dirac delta between any two points in the manifold, or at least a gaussian distribution between any two points even in the limit where x approaches y.

So what I'm thinking is that f is something like
$$f(\alpha ) = \int_x^\alpha {\psi (\alpha ')d\alpha '}$$
so that f(x)=0 and f(y)=1. I suppose that ${\psi (\alpha )}$ could have very large positive and negative values as long as the integral is 1 at y.

8) The Dirac delta, or even just a gaussian, can be manipulated into the Feynman Path Integral of quantum mechanics as I've posted many time in PF, for example, here. These manipulations work also for any gaussians because of the Chapman-Kolmogorov equation, which I can show if asked.

The definition in 4) states, "then there is a continuous function f..." So the question is whether there is anything else that specifies what f can be? Does that mean we are free to invent any functions we like from any point to any other point? Or does the definition mean that if we do have such functions for other reasons, then we must have a manifold?

Or, does the definition mean that the function f must be of the same form no matter where the points x and y are? It's interesting to note that the Dirac delta function integrates to 1 no matter what the interval of integration is, $\int_x^y {\delta (\alpha - \beta )d\alpha } = 1$, as long as $\beta$ is between x and y. So we could use the function, $f(\alpha ) = \int_x^\alpha {\delta (\alpha ' - \beta )d\alpha '}$, for any two points x and y so long as $\beta$ is between the x and y of interest. Or maybe the Dirac delta is not a continuous function, in which case I think the gaussian distribution my work the same way except for infinitesimal differences between x and y.

 

[friend]

Since a continuous function, f, must have a continuous domain, and since every point in the topology can be considered a closed set which must be accommodated in definition 4), it seems that a completely regular space is path connected from any point to any other point. Is that enough to specify a manifold? Or do you also need the property of being Hausdorff?
...
The definition in 4) states, "then there is a continuous function f..." So the question is whether there is anything else that specifies what f can be? Does that mean we are free to invent any functions we like from any point to any other point?

So, if f could be any function satisfying f(x)=0 and f(y)=1, then is this specifying a "function space"? And are there functionals that can be identified on that space?