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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?

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1 Answer 1

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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 $d_1$.

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  • $\begingroup$ How we know that there exists such open set $U\in\tau_1$? $\endgroup$
    – Mittal G
    Feb 24, 2020 at 15:16
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    $\begingroup$ 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 $n>N$. $\endgroup$
    – user722227
    Feb 24, 2020 at 15:18
  • $\begingroup$ Any comments on option 3? $\endgroup$
    – Mittal G
    Feb 24, 2020 at 15:22
  • $\begingroup$ 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?). $\endgroup$
    – user722227
    Feb 24, 2020 at 15:25
  • $\begingroup$ Great. Thanks a lot. $\endgroup$
    – Mittal G
    Feb 24, 2020 at 15:31

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