# Thread: Proving a maximum exists

1. ## Proving a maximum exists

Consider an interval from a to b that we want to show is closed. Given that the interval is bounded and non empty, we know it must contain a supremum. Now we want to show that there exists a max, which will necessarily be the supremum prove that half of the interval is closed i.e. (a,b].

Anyone know how to do so?

2. Originally Posted by cgiulz
Consider an interval from a to b that we want to show is closed. Given that the interval is bounded and non empty, we know it must contain a supremum. Now we want to show that there exists a max, which will necessarily be the supremum prove that half of the interval is closed i.e. (a,b].

Anyone know how to do so?
I don't really understand what you want to prove from what, could you post the whole question?

3. Prove. Let F[a,b]-->R be a continuous function on a compact interval then the set of values of {f(x)|a<=x<=b} has a max and a min.

I know there exists a supremum and infinimum, now I need to show there exists a maximum -- which would be one of my endpoints. Then, repeat the process for a minimum.

4. Okay, you already know the image is bounded so now take a sequence $(f(x_i))_{i\in \mathbb{N} }$ such that $f(x_i) \rightarrow s=sup(f([a,b]))$, this induces a sequence $(x_i) \subset [a,b]$. By Bolzano Weierstrass this has a subsequence $(x_{i_k})$ such that $(x_{i_k}) \rightarrow x \in [a,b]$. Since $f$ is continous $f(x_{i_k})\rightarrow f(x)$ and also $f(x_{i_k}) \rightarrow s$ so $s=f(x)$

5. Can anyway please generalize this property for me? My teacher hinted that it would be on our midterm, but not in the context of continuity..? He just said he would give us a set and we would know that a supremum existed, and that we would have to show a max existed. Ideas anyone?

Thanks.

6. The proof I gave you holds for any function from a sequentially compact (or just compact) metric sapce to the real numbers.

7. Originally Posted by cgiulz
Can anyway please generalize this property for me? My teacher hinted that it would be on our midterm, but not in the context of continuity..? He just said he would give us a set and we would know that a supremum existed, and that we would have to show a max existed. Ideas anyone?

Thanks.
Just let $f(x)=x$.

8. Just to clear things up this makes sense right?

Given a bounded non-empty set $S$ then $\exists$ a $\sup S$ by the completeness property.

Now take a sequence $(a_{n}) \subset S$ such that $(a_{n}) \rightarrow s = \sup S.$ Then by Bolzano Weierstrass there exists a subsequence $(a_{n_i}) \rightarrow m \in S.$ But $(a_{n})$ converges so by the subsequence theorem we have

$(a_{n_i}) \rightarrow m = s.$

Thanks everyone.