Let $f_n : [ a , b ] \longrightarrow R$ satisfy $\mid f_n(x) \mid \leq 1$ for all $x$ and $n$ . Show that there is a subsequence $(f_{nk} )$ such that $\lim_{k\rightarrow \infty}f_{nk}$ exists for each rational $x$ in $[ a , b ]$. Use a diagonalization argument.
I suppose I am to use the fact that the rationals are countable in the given closed interval. I also have in mind the tabular analogy where $f_n$ are the row ids ( thus it goes to infinity) and the rationals are the columns (countable number of them)but finding it difficult to get that subsequence such that the given limit exist.
Enumerate rationals $\{q_j\}_j \subset [a,b]$. Start with $q_1$. Since $|f_n(q_1)| \le 1$ you can find subsequence $\{n_k^{(1)}\}$ such that $f_{n_k^{(1)}}(q_1) \to v_1$, where $v_1 \in [-1,1]$ is some number (we used the fact that from bounded sequence in $\mathbb R^d$ we can extract converging subsequence (Bolzano - Weierstrass)). Now, from $\{n_k^{(1)}\}$ by the same reasoning we can extract subsequence $\{n_k^{(2)}\}$ such that $f_{n_k^{(2)}}(q_2) \to v_2 \in [-1,1]$. Repeat the same for every rational $q_j$ and finally take $n_k = n_k^{(k)}$. Since for any $j \in \mathbb N_+$ it's a subsequence (starting from index $k \ge j$) of $n_k^{(j)}$ we get $f_{n_k}(q_j) \to v_j$ for any $j \in \mathbb N_+$.