Theorem. pointwise boundedness of $\mathscr{C}$ in a countable set [br723]

    If $\{ f_{n} \}$ is a pointwise bounded sequence of complex functions on a countable set $E$, then $\{ f_{n} \}$ has a subsequence $\{ f_{n_{k}} \}$ such that $\{ f_{n_{k}}(x) \}$ converges for every $x \in E$.

    Proof. sketch [br723A]

      Use diagonal magic.

      Let $E = \{ x_{i} \}$. Since $\{ f_{n}(x_{1}) \}$ is a bounded sequence on $\mathbb{C}$, we can find a subsequence that converges to a point of $\mathbb{C}$, denote it by $S_{1} = \{f_{1, k}(x_{1})\}$.

      Now, given $\{ f_{1, k}(x_{2}) \}$, a bounded sequence on $\mathbb{C}$, we can again construct a subsequence that converges, denote it by $S_{2} = \{ f_{2, k}(x_{2}) \}$.

      Similarly, we do the same for every $x_{i}$, and thus obtain sequences $S_{1}, S_{2}, S_{3}, \dots$, which we represent by the array $$ \begin{align*} &S_{1}: \ f_{1,1} \; f_{1, 2} \; f_{1, 3} \; f_{1, 4} \; \dots \\ &S_{2}: \ f_{2,1} \; f_{2, 2} \; f_{2, 3} \; f_{2, 4} \; \dots \\ &S_{3}: \ f_{3,1} \; f_{3, 2} \; f_{3, 3} \; f_{3, 4} \; \dots \\ &\dots \dots \dots \dots \dots \dots \dots \dots \end{align*} $$ Then, $\{ f_{n,n} \}$ is the desired subsequence of $\{ f_{n} \}$.