How does $\mathscr B$ and $\mathscr B'$ are equivalent?

Question

Let $\mathscr T$ and $\mathscr T'$ be topologies on $X$ generated by $\mathscr B$ and $\mathscr B'$ respectively. By Archimedean property I can prove that $ \forall \epsilon>0$ there exists $N \in\mathbb N$ such that $1/N <ϵ $.$\forall x\in X ,B(X;1/N)\subset B(X;\epsilon)$. So, $\mathscr T'$ is finer than $\mathscr T$ by

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How do I prove $\mathscr T$ is finer than $\mathscr T'$?

Answer 1

Hint

If $B\in\mathscr B'$ then $B\in \mathscr B$ because you can choose $\varepsilon=1/n$.