# Confusing proof: closed sets

1. Oct 24, 2009

### workerant

Suppose f:[a,b]--> R and g:[a,b]-->R. Let T={x:f(x)=g(x)}
Prove that T is closed.

I know that a closed set is one which contains all of its accumulation points. I know that f and g must be uniformly continuous since they have compact domains, that is, closed and bounded domains. Now T is the set for which f(x)=g(x), so it seems pretty obvious that this set is going to be closed, but I don't really know how to actually prove this.

2. Oct 24, 2009

### Dick

Have you tried proving that the set of x where f(x) is NOT equal to g(x) is open? If you can show that set is open then its complement is closed.

3. Oct 24, 2009

### Office_Shredder

Staff Emeritus
Hint: Look at the function h(x)=f(x)-g(x). What is T?

4. Oct 24, 2009

### workerant

Thanks
Hmm...okay...so an open set is one where if x belongs to the set T, then there is a neighborhood Q of x that is contained in the set T.

So if T={x: f(x)=/=g(x)}, then we can find a neighborhood of x of T , namely (x-eps,x+eps) that is contained in T. Well....obviously if the function values are not equal, then such a neighborhood exists since our domain is R, but I'm not sure if this proves it formally...

EDIT: Office Shredder, I assume you are going for a different method than Dick? T in that case would be the set for which h(x)=0.

5. Oct 24, 2009

### Office_Shredder

Staff Emeritus
So the complement of T is the set of x so that h(x) =/= 0. It should be easier to think about how to prove that is open (since h(x) is continuous)

6. Oct 24, 2009

### Dick

That's not much of a proof, formal or not. Put delta=|f(x0)-g(x0)|/2. Use the definition of continuity. Office Shredder does have a more direct way of doing it. If you know the theorem he is referring to.

7. Oct 24, 2009

### workerant

Well, then for the complement, there is a neighborhood (x-eps,x+eps) that is contained in T since there are infinitely many points on our domain such that h(x)=/=0 because of the fact that h is continuous.

8. Oct 24, 2009

### Dick

That's an even worse proof than the last one. The last one was not a proof, this one is actually false.

9. Oct 24, 2009

### workerant

I wrote that before I saw the other post you made to set delta equal and use definition.