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If $\{a_n\}$ and $\{b_n\}$ are Cauchy, then $\{a_n + b_n\}$ is Cauchy.


Proof:

$|a_{m_1}-a_{n_1}|\lt \epsilon_1$ and $|b_{m_2} - b_{n_2}|\lt \epsilon_2$

Then take $m_3=\max(m_1,m_2),n_3=\max(n_1,n_2)$

Then $|a_{m_3}+b_{m_3} - a_{n_3}-b_{n_3}|\leq |a_{m_3}-a_{n_3}|+|b_{m_3}-b_{n_3}|\lt2\epsilon$

Now I am unsure how to progress. It would work if my original cauchy sequences were less than $\frac{\epsilon}{2}$, but I don't understand how I would obtain this. Thanks

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  • $\begingroup$ Usually, if we can get $< 2 \epsilon$, we say GOOD ENOUGH! since, in most cases, we could have just said "let such and such so that it is less than $\epsilon / 2$" rather than "let such and such so that it is less than $\epsilon$." $\endgroup$ – MCT Mar 24 '15 at 1:59
  • $\begingroup$ @Soke That's why I like to imagine that I'm working with infinitesimals in calculus, rather than formal, epsilon-delta constructions. If we get it $<2\epsilon$, that's OK, because $2\epsilon$ is also an infinitesimal. $\endgroup$ – Akiva Weinberger Mar 24 '15 at 2:21
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Fix $\epsilon > 0$.

$a_n$ and $b_n$ are Cauchy, so $\exists N_1, N_2$ such that $|a_m - a_n| < \epsilon/2$ when $m, n > N_1$ and $|b_m - b_n| < \epsilon/2$ when $m, n > N_2$.

Choose $N = \max(N_1, N_2)$, then we have

$$|(a_m + b_m) - (a_n + b_n)| \leq |a_m - a_n| + |b_m - b_n| < \epsilon/2 + \epsilon/2 = \epsilon.$$

whenever $m, n > N$, so $\{a_n + b_n\}$ is Cauchy.

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First of all you've made a mistake: you need to introduce $N_1$ and $N_2$ so that for any $m_1,n_1 \geq N_1$ you have the property and similar for the other one.

Having fixed that, if you have $|a_m + b_m - a_n - b_n| < 2 \varepsilon$ for $m,n \geq N$, then you are technically done, since $2 \varepsilon$ can be made arbitrarily small by making $\varepsilon$ arbitrarily small.

Typically for aesthetics, you let $\varepsilon$ be the small number for the converging quantity of interest and then choose new small numbers for what contributes to it from there. Here you can take $\varepsilon$ to be your small number for $a_n+b_n$ and then choose $N_1,N_2$ so that you get $\varepsilon/2$-closeness for $a_n$ and $b_n$ respectively. This is valid because the definition that $a_n$ and $b_n$ are Cauchy says you can pick whatever "$\varepsilon$" you want, so you can in particular pick it to be $\varepsilon/2$ (where $\varepsilon$ was already specified).

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You can choose $\epsilon_1$ and $\epsilon_2$ to be whatever you want, so choose them to be $\frac{\epsilon}{2}$. Choosing a value for $\epsilon_1$ determines a value for $m_1$ and $n_1$, but you can start with any value for $\epsilon_1$ ( and similarly for $\epsilon_2$ ).

The statement that ${a_n}$ is Cauchy is that for any $\epsilon$ there exists an $M$ and $N$ such that for any $n>N$ and $m>M$, we have $|a_m - a_n| < \epsilon$. The quantifiers there are important: $M$ and $N$ depend on the choice for $\epsilon$. But once you know the sequence is Cauchy, you can choose any $\epsilon$, and that will give you a $M$ and $N$. So, typically in these proofs, when you're doing them, you start with what you've done, you get to the end and look at the bound you found ( $\epsilon_1 + \epsilon_2$ ), and then figure out what you needed to start with, for example that $\epsilon_1 < \frac{\epsilon}{2}$ and $\epsilon_2 < \frac{\epsilon}{2}$

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  • $\begingroup$ I thought that might be the case, but why? $\endgroup$ – user142198 Mar 24 '15 at 1:27
  • $\begingroup$ i'll add some more comments to the answer. $\endgroup$ – Callus Mar 24 '15 at 1:28
  • $\begingroup$ Because they are arbitrary real numbers $> 0$. $\endgroup$ – glebovg Mar 24 '15 at 1:29
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You can let $\epsilon' = 2\epsilon$ to complete the proof. However, since in NA people (for whatever reason) insist on using $\epsilon$ at the end of proofs, you could also complete the proof by saying something like "Lastly, interchanging $\epsilon$ and $\epsilon'$ completes the proof."

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