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Given: $c\in \mathbb{R} $ and $ {(a_n})$ sequence and l$\in\mathbb{R}$ and

$\forall \,\, \epsilon>0\,\,\exists\ n_0\in\mathbb{N}$ such that $\,\,\forall n>n_0$, $|a_n-l|

prove that $\forall \,\, \epsilon>0\,$,$\,\exists\ n_0\in\mathbb{N}$ such that $\forall \ n>n_0 \,\,|a_n-l|<\epsilon$

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    that is how it was proven - but now I understand - I begin a fresh proof from scratch choose an $\epsilon$ and then I show that there exist $n_0$ due to the given statement. thanks :)2012-03-27

1 Answers 1

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It's important to realize that the epsilons in the two statements:

$\ \ \ \ $1) $\forall \,\, \epsilon>0\,\,\exists\ n_0\in\mathbb{N}$ such that $\,\,\forall n>n_0$, $|a_n-l|.

and

$\ \ \ \ $2) $\forall \,\, \epsilon>0\,$,$\,\exists\ n_0\in\mathbb{N}$ such that $\forall \ n>n_0 \,\,|a_n-l|<\epsilon$.

are not necessarily the same. You could use any symbol you like in either of the statements to play the role of $\epsilon$. It is therefore valid and would perhaps makes things more clear to reformulate the statements as

$\ \ \ \ $1') \forall \,\, \epsilon'>0\,\,\exists\ n_0\in\mathbb{N} such that $\,\,\forall n>n_0$, |a_n-l|.

and

$\ \ \ \ $2') $\forall \,\, \epsilon>0\,$,$\,\exists\ n_0\in\mathbb{N}$ such that $\forall \ n>n_0 \,\,|a_n-l|<\epsilon$.

You need to show that 2') holds if 1') holds.

Hint:

Note that in order for 1') to hold, we must have $c>0$. Thus, given any $\epsilon>0$, you can, by choosing \epsilon' appropriately, make c\epsilon'=\epsilon.


Solution:

We fix a value of $\epsilon >0$. Then if we set \epsilon'=\epsilon/c we have \epsilon'> 0 (recall that $c$ must be positive). We now appeal to 1') to obtain an $n_0$, so that $\forall \ n>n_0 $: |a_n-l| and we are done.

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    exactly, it took me some time to comprehend it, but after the first answer I got it. Thank you very much for the more detailed solution.2012-03-27