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 (-A). Then A. So -xinf(A). This implies that x-inf(A).
So -inf(A) is an upper bound of -A.
Now let y be any upper bound of -A.
Let xA. Then (-x)(-A). Because y is an upper bound of -A, -xy. This implies that x-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 -yinf(A), which implies that y-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).
If $\displaystyle \lambda = \inf (A)$ then $\displaystyle \left( {\forall x \in - A} \right)\left[ {\; - x \in A \Rightarrow \lambda \; \leqslant - x\; \Rightarrow \; - \lambda \geqslant x} \right]$.
That means that $\displaystyle -A$ is bounded above so $\displaystyle \left( {\exists \delta } \right)\left[ {\delta = \sup ( - A)} \right]\;\& \;\delta \leqslant - \lambda $.
Now suppose that $\displaystyle \delta < - \lambda $ then it follows that $\displaystyle - \delta > \lambda \, \Rightarrow \,\left( {\exists b \in A} \right)\left[ { - \delta > b \geqslant \lambda } \right]$.
BUT this means that $\displaystyle \delta < - b\;\& \, - b \in - A$ : CONTRADICTION.
Let inf(A)=u.
Let $\displaystyle y\in(-A)\Longrightarrow -y\in A\Longrightarrow u\leq -y\Longrightarrow y\leq -u$====> -A is bounded from above by -u.
So if sup(-A)=v$\displaystyle \Longrightarrow v\leq -u\Longrightarrow u\leq -v$......................(1)
Let $\displaystyle - y\in A\Longrightarrow y\in (-A)\Longrightarrow y\leq v\Longrightarrow -v\leq -y$ ====> -v is a lower bound of A.
And since inf(A)=u $\displaystyle \Longrightarrow -v\leq u$...............(2)
Combining (1) and (2) we get u=-v or inf(A)=-sup(-A)