Question:
Let S = {${a_1, a_2, a_3, … a_n}$} be a finite nonempty set of real numbers.
(a) show that S is bounded
(b) Show that lub S and glb S are elements of S
for (a) I see an attempt to answer it was made here: Prove that a finite non empty set of real numbers in bounded .
But the logic is escaping me. Why is M necessarily larger than $a_i$ in the posted example?
(b) I don't know where to begin for this one… I know that if S is bounded then there must be a lub and glb. But why would they necessarily be elements of S? I didn't think a lub or glb had to be an element in a set? Am I misunderstanding?
Best Answer
For part (a), note that for each $i$, $a_i\leq|a_i|$. Also, adding more positive numbers to $|a_i|$ simply makes the number greater. So we get that for each $i$, $a_i\leq|a_i|\leq|a_1|+\cdots+|a_i|+\cdots+|a_n|$. Hence, the set is bounded.
For part (b), note that if a set contains its maximum, then the least upper bound of that set is its maximum. To prove this, let $S$ be a set bounded above that contains its maximum. Then we know it has a supremum $\sup S$, which satisfies the properties that $\sup S$ is an upper bound for $S$, and that any upper bound of $S$ is greater than or equal to $\sup S$. Now note that the maximum of the set $S$ is an upper bound for $S$, since any element of $S$ is less than or equal to the maximum of the set $S$. What can you conclude?