Why do we pick $\epsilon=1$ when trying to prove that every convergent sequence is bounded? So I'm looking to the proof for the theorem "every convergent sequence is bounded". I was studying from both Lima's book and Rudin's book. In their proofs, both use $\epsilon=1 $. The proof is:
Suppose $||a_k||\rightarrow L$. Put $\epsilon=1$. Then, there exists an integer $n>N$ such that 
$$
||a_k-L||<1
$$
We know that $||a_k-L||\geq ||a_k||-||L|| $. Therefore,
$$
||a_k||\leq 1+||L||
$$
For all $n>N$. Now,consider
$$
r=max\{||a_1||,\dots,||a_N||\}
$$
Then, for $k\in \{1,\dots,N\}$ $r\geq ||a_k||$. Finally, it follows that if we pick $r'=max\{1+||L||,r\}$ we have $||a_k||\leq r'$. 
So, there it comes the doubt: why do they pick  $\epsilon=1$. Aren't we supposed to prove for any $\epsilon$? 
 A: 
why do they pick $\epsilon=1$. Aren't we supposed to prove for any $\epsilon$?

Prove what?  If we want to show that $\lim_{n\to\infty} a_n$ exists, then yes, something needs to be shown true for every $\epsilon > 0$.
In this case, though, we are given that $\lim_{n\to\infty} a_n$ exists.  So we can use the “for all $\epsilon > 0$, there exists ...” property to our advantage. 
As for “why set $\epsilon = 1$?”, try a different $\epsilon$ and you will see that the proof still works. You'll still be able to show that the sequence is bounded.  It's just a matter of style which positive number is the most generic, and 1 seems like a good choice.
A: It's perhaps psychological a bit. $1$ is not particularly small, and the choice makes the point that it's not important to choose a small $\epsilon$ -- any $\epsilon > 0$ works. So why not choose the first positive number that we can think of. 
A: The question is not why epsilon is chosen as 1. The question is "why not leave epsilon as epsilon" in the proof, instead of actually picking a value for it. Yes, it works for every epsilon, so we can just leave it as epsilon. Why not?
