Re: Proof of Inf S= -sup(-S)

If S is not bounded below, then -S is not bounded above. What do we say inf S and sup(-S) are in this case?

- Hollywood

Re: Proof of Inf S= -sup(-S)

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**hollywood** If S is not bounded below, then -S is not bounded above. What do we say inf S and sup(-S) are in this case?

- Hollywood

Yes, I understand that if S is not below then inf S=-infinity and if -S is not above, then sup -S = infinity. But is it really just a matter of saying then since inf S = -infinity and sup -S= infinity then -sup -S is -infinity also?

Or am I trying to make it harder than it has to be?

Re: Proof of Inf S= -sup(-S)

Yes, that's all it is.

To be 100% technically correct, the problem statement would have to be "Inf S = -Sup(-S) when they are finite and Inf S = if and only if Sup(-S)= " since the expression is undefined. But that's a technicality - no one thinks is anything other than .

The final case is easy now, right?

- Hollywood