A tough problem

1. Jun 1, 2005

Hurkyl

Staff Emeritus
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.

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.

2. Jun 1, 2005

mathwonk

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
3. Jun 2, 2005

Hurkyl

Staff Emeritus
Bleh. 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
4. Jun 2, 2005

matt grime

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.

5. Jun 2, 2005

mathwonk

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!

6. Jun 2, 2005

Hurkyl

Staff Emeritus
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.

7. Jun 2, 2005

Hurkyl

Staff Emeritus
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.

8. Jun 2, 2005

mathwonk

oops. i knew if it was that easy you would have already done it.

9. Jun 3, 2005

Hurkyl

Staff Emeritus
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.

10. Jun 3, 2005

mathwonk

good work!

11. Jun 3, 2005

Hurkyl

Staff Emeritus
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!

12. Jun 4, 2005

matt grime

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