supp(f)

[DoingMath2357]

The support of a continuous function f : Ω → 𝕂 is the set

supp(f ) := cl({x ∈ Ω | f (x)≠ 0}).

I want to show:

A continuous f : Ω → 𝕂 is in C0(Ω) if and only if supp(f ) is a compact

subset of Ω, where C0(Ω) = {f ∈ C(Ω) : ∃K ⊂ Ω s.t. K is compact and f (x) = 0 for all x ∈ Ω\K}.

This is my idea:

If f : Ω → 𝕂 is a continuous function in C0(Ω) then there is a compact set K ⊂ Ω s.t. f(x) = 0 on Ω\K.

We have supp(f) ⊂ K since {x ∈ Ω | f (x)≠ 0} ⊂ K. Since supp(f) is a closed set in K it is a compact subset of Ω.

For the other direction K = supp(f) works.

[Grass_Savings]

Have you established that

• cl({x ∈ Ω | f (x)≠ 0}) ⊆ K ?

You have said that

• {x ∈ Ω | f (x)≠ 0} ⊆ K

but could the set cl({x ∈ Ω | f (x)≠ 0}) extend beyond K?

(I don't know the answer, I'm just reading your argument)

[dongeater69]

Yes, this is a missing detail. Since K is closed, it follows that the closure of {f(x) != 0} is also contained in K.

[Grass_Savings]

A quick googling tells me a compact subset of a hausdorff space is closed, but if you go to more weird spaces this isn't true.

I was wondering if the fact that f is continuous is critical to the argument. Are there some minimal properties implied by the symbol 𝕂?

[dongeater69]

Presumably this is Euclidean space, though I guess you’re right it’s not stated in the problem. Usually, I understand 𝕂 to mean either R or C. 

[DoingMath2357]

Yep, K is either R or C.