To prove completeness in finer topology?

Question

I encountered an argument which goes like this,

We want to prove topological space ($X,O_1$) is complete(every cauchy sequence converges). Let ($X,O_2$) be topology which is finer(strong) than ($X,O_1$).

To prove ($X,O_1$) is complete, it suffices to prove the completeness of ($X,O_2$).

But I'm confused. If I could prove in coaser(Wearker) topology, completeness follows in more fine topology, and I can say nothing if I work in finer topology,I believe.

Is this kind of argument wrong ? Or am I missing something?

Thank you for your help.

Answer 1

By definition, a sequence $\{a_n:n\in\mathbb{N}\}$ converges to a point $b$ if for every open set $U$ there is $k\in\mathbb{N}$ such that the set $\{a_n:n\geq k\}$ is contained in $U$. So, since in a coarser topology there are fewer open sets to check, whenever a sequence is convergent in a topology $(X,\mathcal{O}_2)$ it would also be convergent in a weaker topology $(X,\mathcal{O}_1)$.

Also, since the definition of completeness (for metric spaces) is that every Cauchy sequence converges, then proving that a finer metric space $(X,d_2)$ is complete suffices to show that a coarser metric space $(X,d_1)$ (same set) is complete.