Let A be a nonempty subset of real numbers which is bounded below. Let -A={-x|x is an element of A). Prove that: inf(A)=-sup(-A).

Any help would be greatly appreciated. Thank you.

decohen

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- Feb 18th 2009, 08:25 AMdecohen@purdue.eduanalysis : infimum/supremum proof
Let A be a nonempty subset of real numbers which is bounded below. Let -A={-x|x is an element of A). Prove that: inf(A)=-sup(-A).

Any help would be greatly appreciated. Thank you.

decohen - Feb 18th 2009, 09:29 AMHalmos Rules
Let http://latex.codecogs.com/gif.latex?x\epsilon(-A). Then http://latex.codecogs.com/gif.latex?(-x)\epsilonA. So -xhttp://latex.codecogs.com/gif.latex?\geqinf(A). This implies that xhttp://latex.codecogs.com/gif.latex?\leq-inf(A).

So -inf(A) is an upper bound of -A.

Now let y be any upper bound of -A.

Let xhttp://latex.codecogs.com/gif.latex?\epsilonA. Then (-x)http://latex.codecogs.com/gif.latex?\epsilon(-A). Because y is an upper bound of -A, -xhttp://latex.codecogs.com/gif.latex?\leqy. This implies that xhttp://latex.codecogs.com/gif.latex?\geq-y. This means that -y is a lower bound of A. Now, since inf(A) is the greatest lower bound of A, it follows that -yhttp://latex.codecogs.com/gif.latex?\leqinf(A), which implies that yhttp://latex.codecogs.com/gif.latex?\geq-inf(A).

In this way we have proved:

1) -inf(A) is an upper bound of -A

2) -inf(A) is smaller than any other upper bound of -A

This means, by definition, that sup(-A)= -inf(A). - Feb 18th 2009, 09:39 AMPlato
If then .

That means that is bounded above so .

Now suppose that then it follows that .

BUT this means that : CONTRADICTION. - Feb 23rd 2009, 08:19 PMbenes