In the proof of the theorem which states that every convergent sequence is bounded, it's chosen $\varepsilon=1>0$, but the proof works for every $\varepsilon>0$ right?
Couldn't the proof be like that:
Suppose $a_n\rightarrow a$, and $\varepsilon>0$ then there is a $n_0\in\mathbb{N}$ such that
$\left | a_n-a \right |<\varepsilon, \forall n\geq n_0$
$\left | a_n \right |=\left|a_n-a +a \right| \leq \left|a_n-a \right|+\left| a\right|<\varepsilon+\left|a \right|$
and let $M=max\left \{ a_1,a_2,...,n_{0-1} \right \}$
For $n\geq n_0$: $\left| a_n \right|<\varepsilon+\left| a\right|$
For $n<n_0$: $\left| a_n\right|<M$
Do we have to specify the value of e since the proof works for every $\varepsilon$? And if so why?