If and , then every element of is an upper bound for the set .

So which is a compact set, and so closed and bounded. Consequently there exists a least upper bound, and and so is a set of upper bounds for .

Is this correct?

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- Sep 27th 2007, 08:48 PMshilz222Compactness: over the top?
If and , then every element of is an upper bound for the set .

So which is a compact set, and so closed and bounded. Consequently there exists a least upper bound, and and so is a set of upper bounds for .

Is this correct? - Sep 27th 2007, 09:14 PMJhevon
- Sep 27th 2007, 09:15 PMshilz222
Oh ok, thanks. I just thought it would sort of nifty and neat.

- Sep 27th 2007, 09:19 PMJhevon
yes, it is nifty. i just learnt about compactness the other day myself, cool stuff. but sometimes we want to be simple and economical. using compactness is fine, your proof will stand out from the rest of the pack. but it depends on what type of professor you have that will determine if that is good or bad

- Sep 29th 2007, 05:15 PMThePerfectHacker
Show that the infimum of T is larger then the supremum of S. Now since T is closed it contains it own infimum.