Proof that infA=−sup(−A)\inf A = -\sup(-A)

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 αx for all xA. Therefore, we know that αx for all xA. Therefore we know that α is an upper bound of A.     

Now let b be the upper bound of A. There exists bxbx for all xA. Hence,
bααbα=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.

aA,xR such that xaxax is an upper bound for A.

Here, x is an arbitrary lower bound for A.

By axiom of completeness, yR such that y=sup(A),i.e. ayx , aA,
which implies that xya.

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

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