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Let $ \epsilon $ >0. show that if $(a_n)_{n=1}$ and $(b_n)_{n=1}$ are eventually $\epsilon$-close sequences, then $(a_n)_{n=1}$ is bounded iff $(b_n)_{n=1}$ is bounded.

proof;

We can choose $M \geq 0$ such that $|a_i| \leq M$ for all $ i \geq 1$.

and we can choose $\epsilon > 0$ and $ N \geq 0$ such that $|a_n - b_n|< \epsilon$ for all n $\geq$N.

we have,

$|b_i| = |b_i - a_i + a_i| \leq |b_i-a_i| + |a_i| \leq \epsilon + M $

Hence $b_i$ is bounded. for all i $\geq$ 1.

Have i done good? or just terrible?

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    $\begingroup$ Looks good to me. $\endgroup$ – BaronVT Jan 15 '15 at 17:46
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    $\begingroup$ thank you ever so much! :) $\endgroup$ – user197848 Jan 15 '15 at 17:48
  • $\begingroup$ It is ok, you have just change $a_n$ and $b_n$ because in the question $(b_n)$ is bounded. $\endgroup$ – Janko Bracic Jan 15 '15 at 17:58
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Your proof looks good. I'd just add the following little details:

  • We have $|a_i-b_i|<\epsilon $ for every $\color{red}{i> N}$. Your inequality implies $|b_i|< \epsilon +M$ for every $\color{red}{i> N}$, and so $$|b_n | \leq \max\{|b_1|,\ldots,|b_N|,M+\epsilon\} \qquad \forall \color{red}{n\geq 1}.$$

  • (this is more fo reading convenience) I would start with the sentence: "Suppose that $(a_n)_{n=1}$ is bounded, then there is $M>0$ such that $|a_i|\leq M$ for every $i \geq 1$..."

  • You proved that if $(a_n)_{n=1}$ is bounded then $(b_n)_{n=1}$ is bounded. But you have to show $(a_n)_{n=1}$ is bounded $\color{red}{\text{if and only if}}$ $(b_{n})_{n=1}$ is bounded. So you could add a little sentence at the end of you proof saying that the case "$(b_n)_{n=1}$ bounded" $\implies$ "$(a_n)_{n=1}$ bounded" is treated similarly (by exchanging the roles of $a_n$ and $b_n$ in the proof for example).

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