Show that $\{a_n\}$ is also a Cauchy sequence in $X$

Question

Let $\{b_n\}$ be a Cauchy sequence in a metric space $X$ and let $\{a_n\}$ be another sequence in X such that $d(a_n, b_n)<\frac{1}{n}$ for every $n\in \Bbb N$. Show that $\{a_n\}$ is also a Cauchy sequence in $X$.

For $\epsilon >0$, there exists $\Bbb N$ s.t. $d(b_n, b_m)<\frac{\epsilon}{mn}$ for $m,n>\Bbb N$

Now consider $d(a_n, a_m)\leq d(a_n, b_n)+d(b_n, b_m)+d(b_m, a_m) \lt \frac{1}{n} + \frac{\epsilon}{mn} + \frac{1}{m} = \frac{m+n+\epsilon}{mn} \lt \epsilon$

$\therefore$ $\{a_n\}$ is Cauchy in $X$.

I have struggled with this proof. I am not sure if what I have is correct. Thank you in advance!

Answer 1

The last inequality is incorrect. Note that $$\frac{m + n + \varepsilon}{mn} \ge \frac{m + n}{mn},$$ so unless you have $m$ and $n$ secretly depending on $\varepsilon$ (hint, hint!), you cannot guarantee that $\frac{m + n}{mn} \le \varepsilon$.

Instead, fix $\varepsilon > 0$, and try to guarantee that $\frac{1}{n}, \frac{1}{m} < \frac{\varepsilon}{3}$. If you choose $N$ large enough, this should be a consequence of $m, n > N$. Then, try a similar triangle inequality argument again.

Answer 2

We have to show, that for any $\epsilon >0$, there exists $N(\epsilon)$, such that for $n,m>N(\epsilon)$: $d(a_m,a_n) <\epsilon$.

So take any $\epsilon > 0$

1) Due, to $(b_n)_{n \in \mathbb N}$ being cauchy, we have, that for $\frac{\epsilon}{3}$, that there exists $M_1(\epsilon)$, such that $d(b_m,b_n) < \frac{\epsilon}{3}$, for $n,m > M_1(\epsilon)$.

2) Due to $\lim_{n \to \infty} \frac{1}{n} = 0$, we have, that for $\frac{\epsilon}{3}$, there exists $M_2(\epsilon)$, such that $\frac{1}{n} < \frac{\epsilon}{3}$, for $n>M_2(\epsilon)$

Now plugging it into your (good idea) and taking $N(\epsilon) = \max\{M_1(\epsilon),M_2(\epsilon)\}$:

$d(a_n,a_m) \leq d(a_n,b_n) + d(b_n,b_m) + d(b_m,a_m) < \frac{1}{n} + \frac{\epsilon}{3} + \frac{1}{m} < \frac{\epsilon}{3} +\frac{\epsilon}{3} +\frac{\epsilon}{3} < \epsilon $