Can we find a divergent Cauchy sequence?

Question

I was wondering if it is possible to build a Cauchy sequence that diverge. That is, a sequence $(x_n)\subseteq X$, for $(X,||\cdot||)$ normed space, such that

(1) $||x_n- x_m||\rightarrow 0$ for $n,m\rightarrow \infty$;

(2) $||x_n||\rightarrow +\infty$.

I think it is not possible. My "rough" idea is that taking $m=O(\log n)$ is pretty the same as fixing m and sending n to infinity. In this way we get a contradiction to Cauchy hypothesis: $$||x_n- x_m||+M\geq||x_n- x_m||+||x_m||\geq ||x_n||\rightarrow +\infty $$ where $M$ is a fixed scalar.

Can you help me formalising it? Or exibit me a counterexample?

Answer 1

Each Cauchy sequence is bounded, so it can not happen that $\|x_n\|\to\infty$.

Let $\epsilon>0$ and $N_0\in\mathbb N$ is such that $\|x_n-x_m\|\leq \epsilon,\,\,\forall n,m>N_0$. Then $\|x_n\|\leq\|x_m\|+\epsilon$. Fix $m>N_0\Rightarrow \|x_n\|\leq \|x_m\|+\epsilon$ for all $n>N_0$

Answer 2

Take simply $X=]0,1[$ and $x_n=\frac{1}{n}$. The sequence $(x_n)$ is a Cauchy sequence that diverge in $X$. If you want a vector space, take $$X=\mathcal C^0([0,1])$$ for the norm $\|f\|=\int_0^1|f(x)|dx$ and $f_n(x)=x^n$. The sequence $(f_n)$ is a Cauchy sequence, but it doesn't converge in $\mathcal C^0([0,1])$.