I asked to prove that a set of real numbers with a supremum not in the set contains a sequence converging to the supremum. I have most of the proof down, but I am not too sure where to bring in the fact that the least upper bound is not in the set.
Proof:
Let $X$ be a set of real numbers with a least upper bound $\alpha$. Suppose $\alpha\notin X$.
Now, because $\alpha$ is the supemum of $X$, for any $\epsilon >0$, there exists an element $x_{n}\in X$ such that $\alpha-\epsilon< x_n$ for all $n\in\mathbb{N}$.
Then, I want to say that because the least upper bound is not in the set, then for $N\in\mathbb{N}$, it follows that $x_n$…..Here is where I am not sure how to contine, but does this make sense to far?
Best Answer
There exists some $x_{1}\in X$ such that $\alpha-1<x_{1}<\alpha$. There exists some $x_{2}\in X$ such that $\max\left\{\alpha-\dfrac{1}{2},x_{1}\right\}<x_{2}<\alpha$. Proceed in this way yields $\max\left\{\alpha-\dfrac{1}{n},x_{n-1}\right\}<x_{n}<\alpha$ with strictly increasing $(x_{n})$ of elements of $X$ and that $x_{n}\rightarrow\alpha$.