# if f:A->B with A closed & bouned, then f(A) is closed & bounded.

• Nov 17th 2009, 06:57 AM
anon2194
if f:A->B with A closed & bouned, then f(A) is closed & bounded.
I have written a valid proof that f(A) is bounded. I don't know how to show that f(B) is also closed.

I am wondering if I can use the Completeness Theorem to argue that there is a supremum and infimum. Then I would need to show that those two numbers reside in A for A to be closed and not open. I don't know how to do that, but am I at least on the right track?

Nevermind, I found out that I am on the right track, and that I just need to use the Maximum-Minimum Theorem.
• Nov 17th 2009, 07:27 AM
tonio
Quote:

Originally Posted by anon2194
I have written a valid proof that f(A) is bounded. I don't know how to show that f(B) is also closed.

I am wondering if I can use the Completeness Theorem to argue that there is a supremum and infimum. Then I would need to show that those two numbers reside in A for A to be closed and not open. I don't know how to do that, but am I at least on the right track?

Nevermind, I found out that I am on the right track, and that I just need to use the Maximum-Minimum Theorem.

And, of course, you must assume f is continuous...(Giggle)

Tonio
• Nov 17th 2009, 07:52 AM
Drexel28
Quote:

Originally Posted by anon2194
I have written a valid proof that f(A) is bounded. I don't know how to show that f(B) is also closed.

I am wondering if I can use the Completeness Theorem to argue that there is a supremum and infimum. Then I would need to show that those two numbers reside in A for A to be closed and not open. I don't know how to do that, but am I at least on the right track?

Nevermind, I found out that I am on the right track, and that I just need to use the Maximum-Minimum Theorem.

I assume since you are discussing the completeness theorem that you are referring to $A,B\subset\mathbb{R}$. How exactly did you prove boundedness without proving closure? Did you use the fact that $A$ is compact?