# How to choose which one is correct?

Suppose that $$\tau_1$$ and $$\tau_2$$ are topologies on $$X$$ induced by the metrics $$d_1$$ and $$d_2$$, respectively, such that $$\tau_1\subseteq \tau_2$$. Then which of the following statement is true?

1. If a sequence converges in $$(X, d_2)$$ then it also converges in $$(X, d_1)$$.

2. If a sequence converges in $$(X, d_1)$$ then it also converges in $$(X, d_2)$$.

3. Every open ball in $$(X, d_1)$$ is open in $$(X, d_2)$$.

My try: Since $$\tau_1\subseteq \tau_2$$, means that every set open in $$\tau_1$$ is also open in $$\tau_2$$. Using this it is easy to show that options 2 and 3 are correct.

But the correct answer is option 1. How to choose it?

As for why 1 is true, let $$S=(a_n)_{n\in\mathbb{N}}$$ be a sequence in $$X$$ that converges w.r.t $$d_2$$. Then if $$U\in\tau_1$$ is open containing the (unique) limit $$x$$ of $$S$$, there exists $$N\in\mathbb{N}$$ with $$a_n\in U$$ when $$n>N$$ (because $$\tau_1\subseteq\tau_2$$). This shows that the sequence converges to $$x$$ w.r.t $$d_1$$.
• How we know that there exists such open set $U\in\tau_1$? Feb 24, 2020 at 15:16
• Well, $X$ is one such open set. The definition of convergence is just that, given arbitrary open $U$ containing $x$, you can find an index $N$ where $a_n\in U$ for all $n>N$. Feb 24, 2020 at 15:18
• Yes. It holds because open balls in $d_1$ are open sets in $\tau_1$ and so are open sets in $\tau_2$. Whether they are actually open balls in $\tau_2$ is a different matter (perhaps that’s what the question means?). Feb 24, 2020 at 15:25