are continuous on A.
We are given that:
,
Let max{ }
Then
i.e. min{ }
Then fg is continuous on A.
This is my proof for the stated theorem. Is there any way I can improve it?
are continuous on A.
We are given that:
,
Let max{ }
Then
i.e. min{ }
Then fg is continuous on A.
This is my proof for the stated theorem. Is there any way I can improve it?
I thought M(x,c) exists just because f and g are defined at x and c. Also, I don't think A has to be closed, bounded, or even an interval for this proof.
And I realized that M(x,c) could be zero... *edit* So I went back and changed the definition of M(x,c). I added the absolute values. Then M(x,c) = 0 iff f(x)=g(c)=0. In that case |f(x)g(x) - f(c)g(c)| = |0 - 0| = 0, and we can just let e_{1} be any in the case of M(x,c) = 0.
Still not following. Given an x and a c, f(x) and g(c) are defined and in R. Then either M = g(c) if f(x) <= g(c), or M = f(x) if g(c) <= f(x). M always exists, so the type of set A is is just restricted by the fact that f and g are continuous on it.
And I am trying to prove that if f,g are continuous at some point c then their product is continuous at that point c. And c is any point in A.
I understand all what you wrote but you are no taking one definite, fixed x but ALL the x's s.t. . The point that is, or should be, fixed is c, the point at which you need to prove continuity! How can you prove that f(x) is bounded no matter what x you choose in ?? you need to use, imo, Weierstrass' Theorem on boundeness of functions in closed intervals.
Tonio
Ok, so if A is closed and bounded, and f is continuous on A, then f is bounded on A.
Then there is an M such that for all x,c in A, f(x)g(c) < M. Choose that as M, so M does not depend on c or x.
Then move into its proper position after . Thanks man, that helped allot (assuming what I just said fixes everything). I had a feeling that moving screwed up something.