LUB (if it exists) of a complete set belongs to that set: Validity

Question

By LUB I mean the least upper bound of the set. And the definition of complete set I am using is that every Cauchy sequence in that set must converge in that set. So by these two assumptions. I cannot seem to get a start on this. I don't even know if this is valid.

Answer 1

Suppose that $S$ is a complete set that has a LUB $x$.

Then for every $n\in\mathbb N$ choose some $s_n\in(x-\frac1n,x]\cap S$.

The set $(x-\frac1n,x]\cap S$ cannot be empty (if it is empty then $x-\frac1{2n}$ is an upper bound of $X$ contradicting that $x$ is its LUB).

Now prove that $(s_n)_n$ is a Cauchy sequence in $S$ hence must converge to some $s\in S$.

Also prove that $s_n$ converges to $x$.

Then apparantly $x=s\in S$.

Answer 2

Let S be set which is complete and Has Least upper Bound say $\alpha$
Case 1: S is finite set.
Then it is easy to show that maximum element of S is LUB of that which contain in S .

Case 2: S is infinite set.
SO for $r>0$ ,$\alpha-r<x<\alpha$ for some $x\in S$ this is true for every r>0.
This is always to true otherwise there is contradiction to lub definition (actually this is the characterisation of Lub In case of infinite set)
Now we can extract sequnce as follows : Chose $x_n\in S $ such that $\alpha-\frac{1}{n}<x_n<\alpha$ with $x_n \neq \alpha$
Actually You have shown Sequnce which is actually convergent to $\alpha $ This is cauchy and hence it limit belong to S
Hence we are done