I'm having trouble with a common proof in real analysis (that I somehow couldn't find a worked example for):
Let $S$ be a non-empty set of real numbers which is bounded above and $\lambda$ is the least upper bound. Prove that for all real numbers $\beta < \lambda $, $\exists \alpha \in S$ such that $\beta < \alpha$
I think I'm taking the wrong approach at this proof or overlooking a helpful theorem somewhere.
I chose to split the proof into two cases:
- $\lambda$ exists in S
- $\lambda$ does not exist in S
The first case completes itself, if $\lambda \in S$, we can say $\alpha = \lambda$ and we've found our valid $\alpha$.
In the second case, if we knew the set was bounded and infinite, then I would know that the least upper bound is an accumulation point (Weierstrass) and therefore gets infinitely close to $\lambda$ and I could construct an $\alpha$ larger than $\beta$ which exists in $S$.
If $S$ is finite will it always contain its least upper bound?
What is the best way to approach this proof?
Yes, if $S$ is finite, it will always contain its least upper bound. This could be proven by induction on the number of elements of $S$, or by a contradiction. Either way, I recommend you try to prove this (and let me know if you get stuck).
The best way to approach this problem would be to assume the conclusion is false: assume there is some real number $\beta<\lambda$ such that for all $\alpha\in S$ we have $\alpha\leq\beta$. Then $\lambda$ is an upper bound of $\beta$ that is strictly less than $\lambda$, contradicting the hypothesis that $\lambda$ is the least upper bound of $S$.