I am reading the proof of Montel's theorem in Stein and Shakarchi Complex Analysis book. I am stuck in page 227, at the start of the proof where we show $\mathcal{F}$ is a normal family. The proof says the following:
Choose a sequence of points $\{w_j\}_{j=1}^{\infty}$ that is dense in $\Omega$. Since $\{f_n\}$ is uniformly bounded, there exists a subsequence $\{f_{n,1}\} = \{f_{1,1}, f_{2,1}, f_{3,1},...\}$ of $\{f_n\}$ such that $f_{n,1} (w_1)$ converges.
What I am stuck here is why does the uniformly bounded condition on the sequence guarantees that we can find a subsequence that converges at $w_1$ and also how does this relate to the fact that $\{w_j\}_{j=1}^{\infty}$ is dense in $\Omega$? Thanks so much for the help or for any hint that helps me understand this better!
If you have a sequence of functions $f_n \colon \Omega \to \mathbb{C}$, evaluating each of these functions at some $w \in \Omega$ gives you a sequence of complex numbers. If the sequence of functions is uniformly bounded (it suffices that it is pointwise bounded), the resulting sequence of complex numbers $z_n = f_n(w)$ is bounded. And then the Bolzano-Weierstraß theorem tells you that you can extract a subsequence that converges [if you can use only the real version, apply it twice, first to get a subsequence for which $x_{n_k} = \operatorname{Re} z_{n_k}$ converges, then to get a subsequence for which $y_{n_{k_m}} = \operatorname{Im} z_{n_{k_m}}$ converges too].
This extraction of convergent subsequences can be repeated for every $w_j$. If we have found a subsequence $(f_{n,k})$ of $(f_n)$ such that $f_{n,k}(w_j)$ converges for all $j \leqslant k$, then since $\bigl(f_{n,k}(w_{k+1})\bigr)$ is a bounded sequence of complex numbers, there is a subsequence $(f_{n,k+1})$ of $(f_{n,k})$ such that $\bigl(f_{n,k+1}(w_{k+1})\bigr)$ converges in $\mathbb{C}$. Since it is a subsequence of $(f_{n,k})$, it follows that $\bigl(f_{n,k+1}(w_j)\bigr)$ converges for all $j \leqslant k+1$. Then the diagonal sequence $(f_{n,n})_{n \in \mathbb{N}}$ has the property that $\bigl(f_{n,n}(w_j)\bigr)$ converges for every $j$, since its tail $(f_{n,n})_{n \geqslant j}$ is a subsequence of $(f_{n,j})$ for every $j$.
Up to here, only the pointwise boundedness of the sequence $(f_n)$ - that means that $\{ f_n(w) : n \in \mathbb{N}\}$ is a bounded subset of $\mathbb{C}$ for every $w \in \Omega$ was used. For the next part, we need that the $f_n$ are holomorphic, and that the sequence is (locally) uniformly bounded. With these conditions, the Cauchy estimates for the derivatives of holomorphic functions show that the sequence $(f_n)$ is equicontinuous.
And now comes the point where we use that $S = \{w_j : j \in \mathbb{N}\}$ is dense in $\Omega$. By the Ascoli-Bourbaki theorem (Bourbaki generalised Ascoli's classical theorem), an equicontinuous sequence of functions that converges at every point of $S$ does in fact converge pointwise on all of $\Omega$, and moreover the convergence is uniform on every compact subset of $\Omega$.
And that is just what we need, from an arbitrary sequence in $\mathcal{F}$ we extracted a subsequence that converges uniformly on every compact $K \subset \Omega$.