# Completely regular spaces

1. Mar 17, 2013

### 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.

2. Mar 17, 2013

### WannabeNewton

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.

3. Mar 18, 2013

### friend

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?

Last edited: Mar 18, 2013
4. May 5, 2013

### 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.

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.

Last edited: May 5, 2013
5. May 9, 2013

### friend

Since a continuous function, f, must have a continous 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?

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.

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.

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.

6. May 15, 2013

### friend

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?