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A tough problem

  1. Jun 1, 2005 #1

    Hurkyl

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    X is a "nice" topological space (insert whatever definition of nice you like). D and E are closed subsets of X, and I have a function f:X-->R which is zero on (D union E) and positive everywhere else.

    The problem is to find functions g and h such that g is zero on D, h is zero on E, and 0 < gh < f everywhere else.


    I've tried working on this problem before, and got nowhere. Been working on it again, and still failing. It's very annoying. :frown:


    More general version: can this be done for any topological space?


    In case you're curious, this problem came from trying to look at the prime spectrum of the ring of continuous, real-valued functions on X. I conjecture to each prime ideal of C(X), the filter comprised of the zero sets of the functions in the prime ideal is an ultrafilter.
     
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  3. Jun 1, 2005 #2

    mathwonk

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    you forgot to say the functions are continuous. as stated it is trivial and any idiot, even i, can solve it immediately.

    i also suppose you, can do it if the space is compact metric. just scale f until the set where it is at most 1/2 consists of points near D and near E. Then construct a function equal to zero on D and equal to 1/2 on the closed complement of the set near D where f is less than 1/2.

    then take the minimum of f or this function to be g. then do same to define h near E. then you are done.

    i don't even tjhikn i need compact for this. or mettrci, just completely regular or whatever that nonsense is.
     
    Last edited: Jun 1, 2005
  4. Jun 2, 2005 #3

    Hurkyl

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    Bleh. :frown: For the general case, I also forgot to state that we require that there exists a function zero precisely on D (and similarly for E)



    I guess I forgot to mention that the interesting case is when D and E are not disjoint -- it sounds like your construction of g would have it zero on some points of E - D, but that's bad.
     
    Last edited: Jun 2, 2005
  5. Jun 2, 2005 #4

    matt grime

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    This doesn't satisfy the constraints precisely, but might be worth a jumping off point:

    Let k=f/2 and let g=h=k^{1/2}

    g is zero on D, and h is zero on E, but obviously g is zero on E and h is zero on D, so it doesn't satistfy the "precisely zero on D" part. Now how about "fudging" g and h slightly, perhaps by adding bump functions, if those were to exist.
     
  6. Jun 2, 2005 #5

    mathwonk

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    take g to be zero precisely on D and positive elsewhere, and h tp be zero on E and positive elsewhere. then multiply them together. and scale f until it is less than 1 everywhere, and ther same for g and h. now taking powers makes evrybody smaller. so at each point x there is a power of (gh)(x) which is less than or eqial to (1/2)f(x).

    hence by compactness, there is one uniform power of (gh) which is less than or equal to f/2. done.

    at least until you remember the rest of your hypotheses!
     
  7. Jun 2, 2005 #6

    Hurkyl

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    The powers idea can fail in two ways (for metric spaces!):

    (1) f might approach zero in increasingly quick ways at different points of the boundary of DUE.

    Each of D and E look like a "polygon" with infinitely many sides, they're attached on one of their sides, and attached to each of the remaining sides there something. (So the things only intersect at the "corners")

    (Okay I suppose you could model it with D being the closed 3rd quadrant, E being the closed 4th quadrant, and on each interval [n, n+1] build a closed equilateral triangle, and let X be the union of all of these things)

    Now, let g(P) be the distance from P to D, and h(P) be the distance from P to E.

    But, on the triangle upon [n, n+1] (n >= 0), let f(P) = d^n where d is the distance from P to E, and similarly on [-n-1, -n].

    So, f approaches some point of DUE like any power of the metric, so no power of (gh) will be everywhere less than f.


    (2) f could approach 0 asymptotically quicker than any power of gh.

    Let X, D, and E be anything, and let g and h be as before. Let f(P) = e^(-1/d2) where d is the distance from P to DUE.

    Now, again, no power of gh will ever be everywhere less than f.
     
  8. Jun 2, 2005 #7

    Hurkyl

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    Incidentally, my goal is to prove (or disprove) each prime ideal of C(X) is contained in a unique maximal ideal... I thought it might help to look at it geometrically, but I'm beginning to think otherwise.
     
  9. Jun 2, 2005 #8

    mathwonk

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    oops. i knew if it was that easy you would have already done it.
     
  10. Jun 3, 2005 #9

    Hurkyl

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    Ah, I figured it out for a metric space -- it's similar to what matt suggested:

    The trick is to find two closed sets D' and E' such that: (& means intersection)

    D' U E' = X
    D' & E = D & E
    D & E' = D & E

    Now, let g' and h' be nonnegative functions whose zero sets are D' and E', then we have:

    g = f + g'
    h = f + h'

    Clearly, g is zero precisely on D and h is zero precisely on E.

    So, we have:

    gh = (f + g')(f + h') = f(f + g' + h') + g' h'

    But, due to the fact that D' U E' = X, we have that g'(x) = 0 or h'(x) = 0 for every x. So, g' h' = 0 (and g' + h' <= 1)

    So, we have gh <= 2f, and I just have to divide one by two to get the functions I actually desire.


    Now, the trick is constructing D' and E'! In a metric space, we can use:

    D' = {P | d(P, D) <= d(P, E)}
    E' = {P | d(P, E) <= d(P, D)}

    I haven't worked out if D' and E' can be found for more general spaces.
     
  11. Jun 3, 2005 #10

    mathwonk

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    good work!
     
  12. Jun 3, 2005 #11

    Hurkyl

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    It feels good to have a partial answer for this -- this particular question has been irritating me for a long time!

    Studying C(X) has really been a fruitful exercise, I think -- it's surprising how much mathematics it has brought together. I'm glad you mentioned in another thread that its maximal spectrum is the Stone-Cech comactification (by the way, just how do you pronounce that?), that's added even more interesting things to look at!
     
  13. Jun 4, 2005 #12

    matt grime

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    Cech is pronounced check, at least it always is when I hear it spoken.
     
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