Proving $\sup A = -\inf(-A)$

Asked Apr 27 '16 at 13:27

Active Apr 27 '16 at 13:37

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In order to prove that

$$\sup A = -\inf(-A)$$

obs: $A$ is limited

I did:

think of $\sup A$ as a number

$$a\le \sup A, \forall a\in A \implies -a\ge -\sup A, \forall a\in A$$ Doesn't that implies that $-\sup A$ is indeed $\inf(-a)?$ I think I should prove that if there is another number that is in the middle, then this number cannot be the sup.

Also, why $A$ has to be bounded?

edited Apr 27 '16 at 13:37

Henno Brandsma

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asked Apr 27 '16 at 13:27

Poperton

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If the $sup$ exists, it has to be bounded above. You don't know whether it's bounded below. What do you mean by a "number that is in the middle?" – Paul Apr 27 '16 at 13:31
It shows that $(-\sup A)$ is a lower bound for $(-A)$. You have to show that it is the greatest lower bound. – MPW Apr 27 '16 at 13:32
What does obs mean? – Henno Brandsma Apr 27 '16 at 13:36

Proving $\sup A = -\inf(-A)$

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