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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?

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  • $\begingroup$ Yes, any positive $e$ would work in the argument. $\endgroup$ – lulu Sep 4 '18 at 11:02
  • $\begingroup$ Specifying the value of $e$ is indeed not mandatory, but it may help understanding what is going on. $\endgroup$ – nicomezi Sep 4 '18 at 11:02
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Yes, every $e>0$ will do and, no, fixing a specific $e$ is not mandatory. But, if you are gong to fix one $e$, $1$ is the natural choice.

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