I have been trying to prove the following statement.
Let $(X, d)$ be a metric space and $\{s_n\}_{n \in \mathbb{N}}$ a sequence of elements of $X$. For $\varepsilon > 0$, consider the two sets \begin{equation*} A_{\varepsilon} := \{ (m, n) \in \mathbb{N} \times \mathbb{N} : d(s_n, s_m) > \varepsilon \} \quad \text{and} \quad B_{\varepsilon} := \{ \min \{n, m\} : (m, n) \in A_{\varepsilon} \}. \end{equation*} The following are equivalent:
- $\forall \, \varepsilon > 0$ the set $B_{\varepsilon}$ is finite, and
- $\forall \, \varepsilon > 0 \quad \exists \, \overline{n} \in \mathbb{N} : \quad \forall \, n, m \geqslant \overline{n} \quad d(s_n, s_m) \leqslant \varepsilon$.
$\boxed{(1) \Rightarrow (2)}$ Let $\varepsilon > 0$ be given. By hypothesis, the set $B_{\varepsilon}$ is finite. Hence, there exists $\overline{n} \in \mathbb{N}$ such that $n < \overline{n}\ \forall \, n \in B_{\varepsilon}$. If $n, m \geqslant \overline{n}$, then we have $d(s_n, s_m) \leqslant \varepsilon$. Otherwise, it would be $\min \{n, m\} \in B_{\varepsilon}$, but this is absurd, because $\min \{n, m \} \geqslant \overline{n}$.
$\boxed{(2) \Rightarrow (1)}$ Let $\varepsilon > 0$ be given. By hypothesis, there exists $\overline{n} \in \mathbb{N}$ such that $d(s_n, s_m) \leqslant \varepsilon\ \forall \, n, m \geqslant \overline{n}$. We can prove that $n < \overline{n}\ \forall \, n \in B_{\varepsilon}$. Let $n$ be a natural number such that $n \geqslant \overline{n}$. For any $m \in \mathbb{N}$, we have $m \geqslant \overline{n}$ or $m < \overline{n}$. If $m \geqslant \overline{n}$, then $d(s_n, s_m) \leqslant \varepsilon$ and $n \notin B_{\varepsilon}$; if $m < \overline{n}$, then $\min \{n, m\} = m$ and $n \notin B_{\varepsilon}$. Therefore, we have $n < \overline{n}\ \forall \, n \in B_{\varepsilon}$.
Is this proof correct?