inf M = -(sup (-M))

ok i tried this:

inf M = a => sup (-M) = -a

- sup (-M) = -(-a) = a

inf M = -sup(-M)

is this good enough proof or am i missing something?

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- Oct 11th 2009, 08:08 AMmetlxprove
inf M = -(sup (-M))

ok i tried this:

inf M = a => sup (-M) = -a

- sup (-M) = -(-a) = a

inf M = -sup(-M)

is this good enough proof or am i missing something? - Oct 11th 2009, 09:35 AMHallsofIvy
This statement assumes what you are asked to prove!

[/quote]- sup (-M) = -(-a) = a

inf M = -sup(-M)

is this good enough proof or am i missing something?[/QUOTE]

No, it isn't. The fact that you haven't actually used an properties of "lim" and "sup" themselves should have been a clue! Use the**definitions**of "inf" and "sup". The inf M is, first of all, a lower bound on M. That is, if $\displaystyle x\in M$, then x> inf M. Can you show that -(sup(-M)) is a lower bound? If [mathj]x\in M[/tex] then $\displaystyle -x\in -M$. Since sup(-M) is an upper bound, we must have -x< sup(-M). Multiply on both sides by -1: x> -sup(-M).

But you still have to prove that -sup(-M) is the**greatest**lowerbound. To do that let m be any lower bound on M.**You will need to show that -m is then an upper bound on -M.**Because "sup(-M)" is the least upper bound, you now have -m> sup(-M). Now multiply on both sides by -1. - Oct 11th 2009, 10:04 AMmetlx
I understand what you're saying but writing it with symbols might be a problem.

We just started with axioms, lower & upper boundaries at uni.. the professor wrote on the blackboard to prove the above and none of us have any idea how.. there's also no proof in the book :/