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$\left\{a_{n}\right\} \rightarrow a$ and $\left\{b_{n}\right\} \rightarrow b$ : show that if $\left\{b_{n}-a_{n}\right\} \rightarrow 0,$ then $a=b$

I am wondering if this is all I needed to do to prove this using some basic properties of sequences?

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    $\begingroup$ It is correct..just note that the limit of a sequence is unique. $\endgroup$ – Marios Gretsas Oct 19 '19 at 18:00
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Your proof is correct. Another way to see it is to observe that $$ \vert a - b \vert = \vert a-a_n + a_n -b_n + b_n -b \vert \le \vert a-a_n \vert + \vert a_n -b_n\vert +\vert b_n - b \vert, $$ where we used the triangle inequality. Fix a $\varepsilon >0$, and choose $N$ so that whenever $n\ge N$, we have that each of $ \vert a-a_n \vert $, $\vert a_n -b_n\vert$, and $\vert b_n - b \vert$ are less than $\varepsilon / 3$. (You may want to convince yourself that we can always find such an $N$ given what we know about the sequences $\{a_n\}$, $\{b_n\}$ and $\{a_n - b_n\}$.)

Choosing $n\ge N$, we see that $\vert a - b\vert < \varepsilon$. This is true for every $\varepsilon > 0$, and so we must have that $\vert a - b \vert = 0$.

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Have you proven that if $k_n \to m$ and $j_n \to l$ then

$(k_n+j_n) \to m+l$.

$vk_n \to vk_n$

$k_nj_n \to ml$

If $l\ne 0$ then $\frac {k_n}{l_n}\to \frac ml$.

If so $b_n\to b$ and $a_n + (b_n -a_n)\to a + 0$ so ....

but if you have not proven that. Well, just do it directly.

For any $\epsilon > 0$ there are $N_1, N_2, N_2$ so that $n > N_1$ implies $|a_n - a| < \frac \epsilon 3$ and $n > N_2$ implies $|b_n -b| < \frac \epsilon 3$ and $n > N_3$ implies $|a_n - b_n|<\frac \epsilon 3$. So for $n \ge N$ then $|b-a| \le |b-b_n| + |b_n - a_n| + |a_n - a| < \frac \epsilon 3+\frac \epsilon 3+\frac \epsilon 3=\epsilon$.

So $|a-b| < \epsilon$ for all $\epsilon$.

It's not nothing to do with limits, but if $|a-b| < \epsilon$ for all $\epsilon > 0$ then $|a-b|=0$ and $a=b$. (Pf: $|a-b| < 0$ is impossible. And if $|a-b|= k> 0$ then our hypothesis is that $|a-b| < k =|a-b|$ which is impossible. So that only leaves $|a-b| = 0$ as an option.)

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