# analysis : infimum/supremum proof

• Feb 18th 2009, 07:25 AM
decohen@purdue.edu
analysis : 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, 08:29 AM
Halmos 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, 08:39 AM
Plato
If $\lambda = \inf (A)$ then $\left( {\forall x \in - A} \right)\left[ {\; - x \in A \Rightarrow \lambda \; \leqslant - x\; \Rightarrow \; - \lambda \geqslant x} \right]$.
That means that $-A$ is bounded above so $\left( {\exists \delta } \right)\left[ {\delta = \sup ( - A)} \right]\;\& \;\delta \leqslant - \lambda$.

Now suppose that $\delta < - \lambda$ then it follows that $- \delta > \lambda \, \Rightarrow \,\left( {\exists b \in A} \right)\left[ { - \delta > b \geqslant \lambda } \right]$.
BUT this means that $\delta < - b\;\& \, - b \in - A$ : CONTRADICTION.
• Feb 23rd 2009, 07:19 PM
benes
Quote:

Originally Posted by decohen@purdue.edu
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

Let inf(A)=u.

Let $y\in(-A)\Longrightarrow -y\in A\Longrightarrow u\leq -y\Longrightarrow y\leq -u$====> -A is bounded from above by -u.

So if sup(-A)=v $\Longrightarrow v\leq -u\Longrightarrow u\leq -v$......................(1)

Let $- y\in A\Longrightarrow y\in (-A)\Longrightarrow y\leq v\Longrightarrow -v\leq -y$ ====> -v is a lower bound of A.

And since inf(A)=u $\Longrightarrow -v\leq u$...............(2)

Combining (1) and (2) we get u=-v or inf(A)=-sup(-A)