Proving a maximum exists

**cgiulz** · October 20th 2009, 03:56 PM

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?

**Jose27** · October 20th 2009, 04:00 PM

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?

**cgiulz** · October 20th 2009, 04:13 PM

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.

**Jose27** · October 20th 2009, 04:20 PM

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 $s=f(x)$

**cgiulz** · October 20th 2009, 06:24 PM

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.

**Jose27** · October 20th 2009, 06:36 PM

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

**redsoxfan325** · October 20th 2009, 07:48 PM

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 $f(x)=x$ .

**cgiulz** · October 20th 2009, 09:57 PM

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.

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