Completely regular space and the Dirac measure

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Does a completely regular space imply the Dirac measure. From wikipedia we have the definition:

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.

And the Dirac measure is defined by:

A Dirac measure is a measure on a set X defined for a given xX and any set AX by δx(A) = 0 for x∉A, and δx(A) = 1 for x∈A.

It seems the definition for a completely regular space includes the definition of a Dirac measure. The difference seems to be that the Dirac measure does not involve a continuous function, but it does seem as though δx(A) = f(x), where the set A for the Dirac measure seems to be the same thing as the set F in the completely regular space. Both f(x)=δx(A)=0 if x∉F or x∉A and 1 otherwise.
 
on Phys.org
What really intrigues me is how we can go from the logic of union and intersection involved with the definition of topology to the math of 1 or 0. Both the definition of a completely regular space and the definition of the Dirac measure seem to do this. Going from sets to numbers is a neat trick.

I'm not trying to suggest that a completely regular space is equivalent to the Dirac measure. I'm wondering if the Dirac measure is implied by a completely regular space. Just because the Dirac measure has a larger domain than the completely regular space does not mean that it doesn't imply the Dirac measure for a restricted portion of that domain. Both seem to be defined in a larger space where you can have points or elements inside and outside a set. And a continuous function from 0 to 1 does imply the existence of a discontinuous function that is either 0 or 1. If a function is defined on a domain, then this implies the existence of the values at it end points.
 
micromass said:
What does that sentence even mean?
If a Completely Regular Space is defined for some F, x, y and f, then does that necessarily allow the construction of a Dirac Measure for a similar A and x? If a closed set F exists (as defined in a CRS) in some topology, then can A also be constructed (as defined for a DM) in that same topology? Or in what kind of topology is A guaranteed to exist if F exists?
 
friend said:
If a Completely Regular Space is defined for some F, x, y and f, then does that necessarily allow the construction of a Dirac Measure for a similar A and x? If a closed set F exists (as defined in a CRS) in some topology, then can A also be constructed (as defined for a DM) in that same topology? Or in what kind of topology is A guaranteed to exist if F exists?

A Dirac measure exists on any set, so yes. You really don't need completely regular.
 
What I'm really driving at is if fields with numeric value are necessarily a part of a manifold. We have fields defined on manifolds as added structure for arbitrary reasons. But are there fields automatically included in the definition of a manifold? We seem to have completely regular spaces as part of the definition of a manifold, and they have functions from 0 to 1. So it seems there are numeric functions automatically part of the definition of a manifold.
 
micromass said:
But why is this so important to you?
Well, let's see. We have SR and GR defined on a manifold. And now manifolds necessarily include fields. I wonder if some of them can be recognized as the quantum fields of SM.
 
friend said:
If a Completely Regular Space is defined for some F, x, y and f, then <Snip>?

But we don't _define_ a complete regular space; we start with a topological space and it is either regular or it is not.
 
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