Transitions from $\frac{1}{n}$ to $\epsilon$ in real analysis proofs I would like to prove the following statement: 

Let $A\subseteq\mathbb{R}$ be nonempty and bounded above, and let $s\in\mathbb{R}$ have the property that for all $n\in\mathbb{N}$, $s+\frac{1}{n}$ is an upper bound for $A$ and $s-\frac{1}{n}$ is not an upper bound for $A$. Prove that $s=\sup A$. 

What I would like to know is if I can use the Archimedean property to "exchange" $\frac{1}{n}$ with some $\epsilon >0$. If I can accomplish this exchange successfully, the rest of the proof is trivial. 
Here is my attempt: By the Archimedean property, for any $y>0$ there exists a number $n\in\mathbb{N}$ such that $y>\frac{1}{n}>0$. Therefore, for any $\epsilon_0>0$ we can choose $\epsilon=\frac{1}{n}$ such that $\epsilon_0>\epsilon>0$. 
I'm not too sure how I can word this to make sense. Basically, how do I convey that the proof using $\frac{1}{n}$ is equivalent to the proof using $\epsilon$? 
 A: I am not sure if this answers your question, but

 Let $s\in\mathbb{R}$. The following are equivalent:
  (i) For all $n\in\mathbb{N}$, $s+\frac1 n$ is an upper bound for $A$, but $s-\frac1 n$ is not an upper bound for $A$.
  (ii) For all $\epsilon>0$, $s+\epsilon$ is an upper bound for $A$, but $s-\epsilon$ is not an upper bound for $A$.
  

$(i)\Rightarrow(ii)$: Let $\epsilon>0$. Then there exists $n$ such that $\frac1 n<\epsilon$. In this case, $s+\frac1 n$ is an upper bound for $A$, by $(i)$, in particular, $s+\epsilon>s+\frac 1n$ is an upper bound as well. Moreover, there exists $y\in A$ such that $s-\frac1 n<y$, since $s-\frac1 n$ is not an upper bound for $A$, by (i). Then $s-\epsilon<s-\frac1 n<y$, hence $s-\epsilon$ is not an upper bound for $A$ as well. This proves (ii).
$(ii)\Rightarrow(i)$: trivial, since we can take $\epsilon=\frac1 n>0$, for any $n\in\mathbb{N}$.
A: Yes you can do that.
Let $n > \frac 1{\epsilon}$.  Then $\frac 1n < \epsilon$.
So for any $p < s$, let $\epsilon= s-p > 0$ then if $n > \frac 1 {\epsilon}$ then $p = s-\epsilon < s-\frac 1n$ so $p$ is not upper bound so $\sup A \ge s$.
For and $r >s$, let $\epsilon = r -s > 0$ and $n > \frac 1{\epsilon}$ so $s + \frac 1n < r$ so $s + \frac 1n$ is an upper bound so $r$ is an upper bound but not the least upper bound.  
So $\sup A \not > s$ so $\sup A \le s$.
So $\sup A = s$.
A: We first show that $s$ is an upper bound for $A$. That is, for any $a\in A$, $a\le s$. 
Suppose this is not the case, then there exists some $a'\in A​$ such that $a' > s​$. Since $s + \frac{1}{n}​$ is an upper bound for any $n$, in particular, we can choose $n'$ such that $\frac{1}{n'} < \frac{a' -s}{2}$. This is possible due to Archimedean property since $a'-s > 0$.
This implies that $s + \frac{a' - s}{2}$ is an upper bound, which  implies that $s + \frac{a' - s}{2} \ge a'$, which implies $s \ge a'$, a contradiction.
You can finish the second part using similar argument.
A: Let's say you want to show that for every $\varepsilon>0$ there exists some $\delta_\varepsilon>0$ such that $\varphi(\delta)<\varepsilon$ for all $\delta\leq \delta_\varepsilon$, where $\varphi(\cdot)$ is an expression you're interested in, which depends on some real input. Now suppose that you're only able to show for every $n\in \mathbb N$ there exists a $\delta_n>0$ such that $\varphi(\delta)<\frac{1}{n}$ for all $\delta\leq\delta_n$. This is in fact enough to show what you want, as you rightly expect. Why? You have the right idea, but since you asked let us give a semi-rigorous argument:
Fix $\varepsilon>0$. By the Archimedean property there exists a $n\in N$ such that $\varepsilon\geq\frac{1}{n}$. Now by assumption you can find some $\delta_n>0$ such that $\varphi(\delta)<\frac{1}{n}$ for all $\delta\leq \delta_n$. Choose $\delta_\varepsilon=\delta_n$. Then for all $\delta\leq\delta_\varepsilon$ we have that $\delta\leq\delta_n$, so $\varphi(\delta)<\frac{1}{n}\leq \varepsilon$.
