Searching for a sequence that shows something isn't a complete metric space [duplicate]

HINT: The function

$$f:\Bbb R\to(-1,1):x\mapsto\frac{x}{1+|x|}$$

is a homeomorphism, and $d(x,y)=|f(x)-f(y)|$ is just the usual Euclidean distance between $f(x)$ and $f(y)$ in $(0,1)$. Find a Cauchy sequence in $(-1,1)$ that does not converge to any point of $(-1,1)$, and use $f$ to ‘pull it back’ to a sequence in $\Bbb R$.

Observe the behaviour of the function $f(x) = \frac{x}{1 + |x|}$. As $x \to \infty$ in $\mathbb{R}$, $f(x)$ will get closer and closer to $1$. Intuitively, the metric will squeeze things together for you as you move to the right on the real line, in the sense that $1000, 1000+1$ are much closer than $1, 1+1$ under this metric.

To exploit this, consider the sequence $a_n = n$. Let $\epsilon > 0$ be given. Then note that for $N$ large enough, you can have $1 - \epsilon < f(a_N) < 1$. Then, observe that for any $a_m, m \ge N$, you will also have $1 - \epsilon < f(a_m) < 1$, which means that if $n, m \ge N$, $f(a_n)$ and $f(a_m)$ lie in the same interval $(1-\epsilon, 1)$, which implies $|f(a_n )- f(a_m)| < \epsilon$. So the sequence is Cauchy. Next, you show that this sequence does not have a limit point, hence does not converge. Then you are done!