Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Completely regular space and the Dirac measure

  1. May 6, 2015 #1
    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 continous 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.
  2. jcsd
  3. May 6, 2015 #2
    In the definition of the completely regular space, the function ##f## has ##X## as domain. The Dirac measure has ##2^X## as domain. There really is no connection between the two. Furthermore, the Dirac measure can be put on any set, you don't even need a topology for it.
  4. May 8, 2015 #3
    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.
  5. May 8, 2015 #4
    What does that sentence even mean?
  6. May 8, 2015 #5
    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?
  7. May 8, 2015 #6
    A Dirac measure exists on any set, so yes. You really don't need completely regular.
  8. May 8, 2015 #7
    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.
  9. May 8, 2015 #8
    Yes, the definition of a (smooth) manifold implies the existence of many (smooth) scalar fields. The essential theorem here is the existence of a (smooth) partition of unity on the manifold.

    But why is this so important to you?
  10. May 8, 2015 #9
    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.
  11. May 8, 2015 #10


    User Avatar
    Science Advisor
    Gold Member

    But we don't _define_ a complete regular space; we start with a topological space and it is either regular or it is not.
  12. May 8, 2015 #11


    User Avatar
    Science Advisor
    Gold Member

    Still, @friend, it seems you are looking maybe for some categorical relationship between , maybe regular topological spaces and continuous functions, with measure spaces with the Dirac measure?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook