Let A be a nonempty subset of real numbers which is bounded below. Let −A be the set of of all numbers −x, where x is in A. Prove that infA=−sup(−A)

So far this is what I have

Let α=inf(A), which allows us to say that α≤x for all x∈A. Therefore, we know that −α≥−x for all x∈−A. Therefore we know that −α is an upper bound of −A.

Now let b be the upper bound of −A. There exists b≥−x⟹−b≤x for all x∈A. Hence,

−b≤α−α≤b−α=−inf(A)=sup(−A)By multiplying −1 on both sides, we get that inf(A)=−sup(−A)

Is my proof correct?

**Answer**

Let A be a nonempty subset of real numbers which is bounded below. Let −A be the set of of all numbers −x, where x is in A. Prove that infA=−sup(−A).

I think that the purpose of this question is to show you why it is not required to include the existence of infimum into the Axiom of Completeness.

Here, i will show the existence of infimum of A.

∀a∈A,∃x∈R such that x≤a⟹−x≥−a⟹−x is an upper bound for −A.

Here, x is an arbitrary lower bound for A.

By axiom of completeness, ∃y∈R such that y=sup(−A),i.e. −a≤y≤−x , ∀a∈A,

which implies that x≤−y≤a.

Here, −y is a lower bound for A and −y is at least any lower bound for A, which means that −y=infA.

Hence, −sup(−A)=infA.

**Attribution***Source : Link , Question Author : user77107 , Answer Author : Little Rookie*