Proof concerning equivalent definition of supremum using limits I believe I have a proof to the following theorem:
Let $S$ be a subset of $\mathbb{R}$ that is non-empty and bounded above.
 $s \in \mathbb{R}$ is the supremum iff $s$ is an upper bound of $S$ and for all $\epsilon>0$ there is an $x \in S$ such that $abs(s-x) < \epsilon$.
now the forward part of the implication is easy because I am given the theorem
if $s=sup(S)$ then for every $\epsilon > 0$ there is an element $x \in S$ such that $s-x < \epsilon$.
However, the reverse part was a little more difficult and I am not sure if I did it right so I wanted to post it here to check.
(by reverse part, I mean: if $s$ is an upper bound of $S$ and for all $\epsilon>0$ there is an $x \in S$ such that $abs(s-x) < \epsilon$ then $s=sup(S)$ ).
here is my proof:
suppose for a contradiction that $s$ is an upper bound of $S$ and for all $\epsilon>0$ there is an $x \in S$ such that $-\epsilon < s-x < \epsilon$, but that $s \neq sup(S)$. That is, there exists a real number $u$ such that for all $x \in S$ : $s \geq u \geq x$. Now pick $u = \epsilon$ and $x = 0$ so we have that $-\epsilon < s < \epsilon$ and $s \geq \epsilon \geq 0$. But that means that $s < \epsilon$ and $s \geq \epsilon$, which is an absurdity. Thus If $s$ is an upper bound of $S$ and for all $\epsilon>0$ there is an $x \in S$ such that $abs(s-x) < \epsilon$ then $s=sup(S)$.
My concern for the proof is with regards to setting $u = \epsilon$ and $x = 0$, am I allowed to do this or should $x $ depend on $\epsilon$? From what I have read about proofs online is that when an existential quantified variable ($x $ in this case) follows a universal quantified variable ($\epsilon$ in this case) then $x $ should depend on $\epsilon$, however how can I guarantee that there is an element $x $ in $S $ that is equal to $\epsilon$ when I know nothing about the set $S $?
 A: Your argument starts off correctly, but you make a small mistake by taking $\epsilon = u$. Note that $\epsilon > 0$ whereas $u$ may or may not be positive. We can complete the proof as follows:
Let $s$ be an upper bound for set $S$ and for any $\epsilon > 0$ there exist an $x \in S$ dependent on $\epsilon$ such that $|x - s| < \epsilon$. If $s \neq \sup S$ then it means there is an upper bound $s'$ of $S$ such that $s' < s$. Therefore for all $x \in S$ we have $x \leq s' < s$. It means that $|x - s| = s - x \geq s - s'$. If the value of $\epsilon$ is taken to be less than $s - s'$ then we get a contradiction. Hence we must have $s = \sup S$.

Update: After going through your argument one more time I found that you were trying to use the inequality $|x - s| < \epsilon$ and somehow derive a contradiction. To that end you chose a specific value $x = 0$. This is not allowed because the value of $x$ depends on $\epsilon$ and can't be chosen arbitrarily. In fact it may happen that $0$ is not a member of $S$ so that $x = 0$ is not allowed in any way. What we really know about the $x$ appearing in the inequality $|x - s| < \epsilon$ is that $x$ is some element in $S$ which depends on $\epsilon$. There is just no way to figure out any specific value of $x$ because we don't know anything about members of $S$ except that they are all bounded above by $s$.
Now let us understand the meaning of inequality $|x - s| < \epsilon$ in detail. The symbol $|x - s|$ represents the distance between $x$ and $s$ on number line. So what this means is that we can find an $x$ whose distance from $s$ is less than $\epsilon$. Since $\epsilon > 0$ is arbitrary it means that we can find $x \in S$ such that its distance from $s$ is as small as we please. This is illustrated in the figure below on the number line.
----------------------$x$--------$s$---------------------------------------------------
--------------------------$x$----$s$---------------------------------------------------
----------------------------$x$--$s$---------------------------------------------------
-----------------------------$x$-$s$---------------------------------------------------
Now consider the argument when $s$ is not the suprememum. In that case we have an upper bound $s' < s$ so that the above figure changes to
------------------------$x$--------$s'$--------$s$---------------------------------------
Since $s'$ is an upper bound therefore the value of $x \in S$ will always be less than or equal to $s'$ so that $x$ will never go to the right of $s'$ on the number line. And since $s$ lies to the right of $s'$ it means that $x$ can never go as close to $s$ as we want. More technically the distance between $x$ and $s$ will always be greater than or equal to distance between $s$ and $s'$. Thus we can take any $\epsilon$ with $0 < \epsilon < s - s'$ and get a contradiction.
While studying analysis (especially introductory parts) it is important to realize that the inequalities have meanings which are best expressed via a number line diagram. Often this fact is not stressed and the student feels that inequalities are just symbol manipulation based on certain rules. Once you start to get the meaning behind such inequalities like $|x - s| < \epsilon$ by putting numbers involved on the number line then such difficulties will be easily overcome.
