Let $S$=Sup $M.$ Prove there exists a sequence ($a_n$)$_{n=1}^{\infty}$ that converges to $S$ such that $a_n$ $\in$ $M$ for all natural numbers $n$.
Is there a way to prove this without assuming the sequence increases and without using the squeeze theorem?
So far, I have this:
By the definition of a supremum of a nonempty set, $M,$ we know that Sup $M$=$S$ is the least upper bound of $M.$ If we have a sequence $a_n$ $\in$ $M$ for all natural numbers $n,$ we can construct $a_n$ $\in S$ such that |$a_n – S$| < $\epsilon$ where $\epsilon$=$1\over n.$ Then -|$1\over n$|$\leq$|$a_n-S$|$\leq$|$1\over n$|.
Since |$a_n-S$| is bounded, we know that there exists a number, $x\in M$ such that $S – \epsilon \leq x < S$ for all $\epsilon$ >0.*
*This last sentence is from a previously proven theorem which states that:
Suppose $M = (a,b)$, and Sup $M=S$. Then for every number $\epsilon$ > 0, there exists a number $x$ in $M$ such that $S – \epsilon \leq x < S.$
I'm not sure if I am on the right track, and was hoping for guidance in solving this proof without using the Squeeze theorem and without assuming the sequence increases.
Best Answer
This statement:
is stated incorrectly. In fact, there can be no such $x$ since it would satisfy $S \le x < S$.
Prior to that you essentially had the right answer. For each $n$ you can find $a_n \in M$ satisfying $S - \frac 1n < a_n \le S$, from which it follows $|a_n - S| < \frac 1n$ for all $n$.
Given $\epsilon > 0$ choose $N$ satisfying $\frac 1N < \epsilon$. Then $$n \ge N \implies |a_n - S| < \frac 1n \le \frac 1N < \epsilon.$$ This is the definition of $a_n \to S$.