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I am trying to understand the proof of the proposition:

Any convergent sequence is bounded.

In my textbook, the author uses the definition of convergence for a sequence $\{a_n\}\to l$ and fixes $\epsilon=1$ so that there is a natural number $N$ such that \begin{align*}n>N&\implies|a_n-l|<1\\&\implies |a_n|<1+|l|.\end{align*} Then $\{a_n\}$ is bounded by $\pm U$ where $U=\max\{|a_1|,|a_2|,|a_3|,\dots,|a_{N-1}|,|a_N|,1+|l|\}$.

What I don't understand about this proof is why do we have to fix $\epsilon=1$? Wouldn't it be enough if we had simply fixed some $\epsilon>0$ and then claimed that \begin{align*}n>N &\implies |a_n-l|<\epsilon\\&\implies |a_n|<\epsilon+|l|.\end{align*}In this case $\{a_n\}$ would be bounded by $\pm M$ where $M=\max\{|a_1|,|a_2|,|a_3|,\dots,|a_{N-1}|,|a_N|,\epsilon+|l|\}$. So, why does the author decide to fix $\epsilon$ to be 1 when it would be enough fix some $\epsilon$. For example, what if I had fixed $\epsilon=\pi$, would the proof be incorrect?

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    $\begingroup$ You are correct. You can use any $\epsilon$. Just choose one. $\endgroup$
    – user99914
    Commented Oct 19, 2013 at 23:09
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    $\begingroup$ One must fix some $\varepsilon$. Whether one chooses $1$ or $\pi$ or $42$ doesn't matter. $\endgroup$
    – Daniel Fischer
    Commented Oct 19, 2013 at 23:10

1 Answer 1

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Yes, you could choose an other positive value of $\epsilon$ and the proof is identical. The only reason for choosing $\epsilon = 1$ is that it's a slightly more convenient choice.

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