What's wrong with this proof? / uniqueness of least upper bound this is a proof by contradiction
let y and  z be least upper bounds of a set A, such that y != z
so, according to a theorem, L - ε < x,for all x in A. where L is the least upper bound and  ε is a positive real number.
so my proof goes like this
according to that theorem, we have
1) y -  ε < x, for all x in A 
2) z -  ε < x, for all x in A
now, by adding the inequalities, we get 
3) y + z < 2x, for all x in A
now by multiplying by 1/2, we get
4) (y + z)/2 < x, for all x in A
(y + z)/2 is the arithmetic mean of y and z, since both y and z are the least upper bound of A, (y + z)/2 must be greater than x, but equation 4 tells us the opposite, so that's the contradiction.
when I showed this to my professor, he said it was wrong, he told me that the arithmetic mean could also be a least upper bound, but honestly, I don't understand, I don't see why the arithmetic mean could also be a least upper bound.
 A: The proof is easier.
Let $y$ and $z$ be two distinct least upper bounds. Assume $y<z$ (otherwise swap them). But as $z$ is least, for any upper bound $y$, we have $z\le y$ !
A: The least upper bound of a set S is some x such that:
$\forall{s \in S}, x \ge s.$
$\forall {x'\lt x}, x'$ is not an upper bound.
For a set S with upper bound $B$:
Define a sequence of nested intervals $I_n=[a_n,b_n]$ as the following:
$a_1=a$, an element in S that is not an upper bound, $b_1=B$
$$\forall{k\in \mathbb N}, a_{k+1}=a_k, b_{k+1}=\frac{a_k+b_k}{2}\;\text{if $\dfrac{a_k+b_k}{2}$ is an upper bound},a_{k+1}=\frac{a_k+b_k}{2}, b_{k+1}=b_k\;\text{if $\dfrac{a_k+b_k}{2}$ is NOT an upper bound} $$
So $lim_{k\to\infty}a_n=lim_{k\to\infty}b_n$ exists , call it $x$.
x is an upper bound of S since $\forall{k \in \mathbb N}, b_k$ is an upper bound of S. Now let us suppose there is some $x'\lt x$ such that $x'$ is also an upper bound.Therefore,
$$\forall{k\in\mathbb N},a_k\lt x' \lt x \lt b_k$$
Then $|b_k-x'|\gt |b_k-x| \;\forall{k\in\mathbb N}$. 
Now by the definition of limits, $\forall{\varepsilon_1,\varepsilon_2 \gt 0}, \exists{N_1,N_2\in\mathbb N}$ such that $n\ge N_1$ implies $|a_n-x|\lt \varepsilon_1, n\ge N_2$ implies $|b_n-x|\lt \varepsilon_2$
Let $N=max\{N_1,N_2\}$. Then by the triangle inequality, $$k\ge N,|b_k-a_k|=|(b_k-x)+(a_k-x)| \le |b_k-x|+|a_k-x|\lt \varepsilon_1+\varepsilon_2:=\varepsilon$$
$x-x'\lt b_k-a_k \;\forall{k\in\mathbb N}$, so if we set $$\varepsilon_1=\varepsilon_2=\frac{x-x'}{2}$$ Then $\exists{N_0 \in \mathbb N}$ such that for $k\ge N_0, b_k-a_k\lt x-x'$, which is a contradiction. Hence no smallest upper bounds than $x$ exist, and therefore $x$ is the least upper bound, and clearly this least upper bound is therefore unique as well. In other words, once one has established the existence of the least upper bound for a set S, it's uniqueness is trivial.
