# Prove that if $\{a_n-b_n\}\to 0$ then $a=b$.

$$\left\{a_{n}\right\} \rightarrow a$$ 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?

• It is correct..just note that the limit of a sequence is unique. – Marios Gretsas Oct 19 '19 at 18:00

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 $$\vert a - b \vert = 0$$.

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.)