If f : U → R is continuous on U, and E ⊂ U is closed and bounded, then(adsbygoogle = window.adsbygoogle || []).push({});

f attains an absolute minimum and maximum on E.

How do you prove this theorem? I asked about this a while ago, but now I have a chance to redo the assignment and I need to fix my proof. I started by proving it for the 1 dimensional case. Here it goes:

for [tex]F:U\subset\Re\rightarrow\Re[/tex]

It is easy to show by continuity that everypoint has an epsilon neighborhood on which the function is bounded. I will only show this if someone asks. so if we consider f on the interval (a,b), then consider the set of x such that f is bounded on (a,x) and x is in (a,b). From what was just said, the set is non-empty. Let c be the largest upper bound of this set. Assume c<b, than f is bounded on (a,c-e), but since it is also bounded on (c-e,c+e) it is bounded on (a,c+e). This contradicts our choice of c, therefore c=b, thus f is bounded on (a,b-e) but also on (b-e,b) and therefore on (a,b).

To show that F achieves its bounds, consider [tex]g(x)=\frac{1}{M-f(x)}[/tex]

where M is either the upper or lower bound of f. Then if f(x) does not equal M on some point in (a,b), then g(x) is continuous on (a,b) and thus, by argument above is bounded on (a,b). This clearly cannot be true therefore f(x)=M for some x in (a,b)

QED

My problem is in trying to extend this to the n-dimensional case. First of all can it be extended? If it can, my problem is how do you use the trick of using contradiction to show that the function is bounded from one boundary point to another one, because the boundary is no longer just 2 points? What I had handed in was that since the function is bounded on an epsilon neighborhood of every point, take the union of such neighborhoods to be a covering of E by open sets, call it K. Than F is bounded on K, since it is a union of sets on which F is bounded. Since E is contained in K, F is therefore bounded on E. Than the same argument used before can be used to show that F attains its bounds.

The problem with this proof is that I am wrong in saying that F is bounded on K because it is a union of sets on which F is bounded. This is because that is not necessarily true if it is a union of infinite sets.

Is there any way to show that a covering of E on which F is bounded can be made by a union of a finite number of sets on which F is bounded? Or is there no way to salvage this proof?

My difficulty here is that we have been taught little to no topology (most of the terminology used in this post i learnt on my own on the internet) and so I need a proof that doesn't depend on too many other theorems. We've basically been given the definitions of open, closed, bounded, compact, continuous and limit, and that's about it.

Thanks for any and all help.

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# A function on a compact set is compact

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