Let S and T be non-empty, bounded sets of real numbers and suppose that s < t for $\forall$s $\in$ S and $\forall$t $\in$ T
Which of the following statements are true:
(i) inf S $\leq$ inf T
(ii) inf S $\leq$ sup T
(iii) sup S $\leq$ sup T
(iv) sup S $\leq$ inf T
So far I have:
(i) inf S $\implies$ inf S $\leq$ x for all x $\in$ S
Since s < t for $\forall$s $\in$ S and $\forall$t $\in$ T, this implies that inf S $\leq$ x for all x $\in$ T therefore inf S is a lower bound for T.
inf T means that if k is another lower bound for T then k $\leq$ inf T.
Let k = inf S (since both are lower bounds for T) therefore inf S $\leq$ inf T
(ii) sup T $\implies$ sup T $\geq$ x for all x $\in$ T
s < t for $\forall$s $\in$ S and $\forall$t $\in$ T $\implies$ sup T is a an upper bound for S
Since inf S $\implies$ inf S $\leq$ for all x $\in$ S, this implies that inf S $\leq$ sup T
Is that going in the right direction?
Since every $t\in T$ is an upper bound of $S$ and since $\sup S$ is the least upper bound of $S$, $\sup S\leqslant t$, for each $t\in T$. But then, by a simliar argument, $\sup S\leqslant\inf T$.
And now, since $\inf S\leqslant\sup S$, $\inf S\leqslant\inf T$ too.
By a similar argument, $\sup S\leqslant\sup T$.
And now, of course we have that $\inf S\leqslant\sup T$.