Proof of Arzela-Ascoli Theorem on Conway's book

Question

CONTEXT

Conway's Functions of One Complex Variable I, page 148, reads

1.23 Arzela-Ascoli Theorem. A set $\mathcal F \subset C(G,\Omega)$ is normal iff the following two conditions are satisfied.

(a) For each $z$ in $G$, $\{f(z) : f\in \mathcal F\}$ has compact closure in $\Omega$.

(b) $\mathcal F$ is equicontinuous at each point of $G$.

Proof. First assume that $\mathcal F$ is normal. Notice that for each $z$ in $G$ the map of $C(G,\Omega) \to \Omega$ defined by $f \mapsto f(z)$ is continuous; since $\mathcal F^-$ is compact its image is compact in $\Omega$ and (a) follows.

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Note that $C(G, \Omega)$ is the set of all continuous functions from $G$ to $\Omega \in \mathbb C$ with the usual metric $\rho$ defined in 1.4, page 143.

$\mathcal F^-$ denotes the closure of $\mathcal F$. A set is normal iff its closure is compact.

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QUESTION

Let $\psi_z$ the map defined by the author in the proof. Since $\psi_z$ is continuous we know that $$\psi_z(\mathcal F^-) \subseteq \left(\psi_z(\mathcal F)\right)^-=\{f(z) : f\in \mathcal F\}^-.\tag{1}\label{1}$$ So, even if continuity and compactness of $\mathcal F^-$ imply compactness of $\psi_z(\mathcal F^-)$, we cannot conclude immediately that the closure of $\{f(z) : f\in \mathcal F\}$ is compact.

Is my way of reasoning correct?

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WHAT I HAVE TRIED

I tried to get to the conclusion with some additional thoughts. I am also asking you if these are correct or erroneous/unnecessary.

First note that $\psi_z$ is surjective. Let $w \in \left(\psi_z(\mathcal F)\right)^-$, so that there exists a sequence $(z_k) \to w$, with $z_k \in \mathcal F$. By definition of $\psi_z$ and by surjectivity, therefore, there exists functions $f_k$ and $f$ such that $f_k(z) = z_k$ and $f(z) = w$, so that now we have $$(f_k(z)) \to (f(z)).$$ Fix $\delta >0$. For large enough $k$ we have $$|f_k(z)-f(z)| < \frac{\delta}3.$$ By continuity of $f_k$ and $f$, there exists a compact set $K$ such that $$|f_k(\zeta)-f_k(z)| < \frac{\delta}3$$ and $$|f(\zeta)-f(z)| < \frac{\delta}3,$$ for all $\zeta$ in $K$, so that $$|f_k(\zeta)-f(\zeta)| \leq |f_k(\zeta)-f_k(z)| + |f_k(z)-f(z)|+|f(\zeta)-f(z)|<\delta.$$ Since $\delta$ was arbitrary, we can choose $\delta$, and thus $k$, so that (Lemma 1.7, page 144) $$\rho(f_k,f) < \varepsilon,$$ for any $\varepsilon >0$. This shows that $f \in \mathcal F^-$, and since $f(z) = w$, we have that $$\psi_z(\mathcal F^-) \supseteq \left(\psi_z(\mathcal F)\right)^-.$$ Together with \eqref{1}, this leads to our conclusion, i.e. that $\psi_z(\mathcal F^-) =\left(\psi_z(\mathcal F)\right)^-$ and that this set is compact.

Answer 1

I haven't yet checked your working, but I believe Conway was justified in stating its compactness as an 'immediate' fact.

Fix a $z\in G$ and let $\phi:C(G,\Omega)\to\Omega$ be the evaluation at $z$. You are given that this function is continuous (I don't think this is true if $G,\Omega$ are arbitrary spaces, so I'd be curious to know what the context is here) so we know $\phi(\overline{\mathcal{F}})$ is compact as $\mathcal{F}$ is precompact (btw, "normal" in topology tends to refer to the condition on a space that disjoint closed sets are separable).

I know that $\phi(\mathcal{F})\subseteq\phi(\overline{\mathcal{F}})$. If $\Omega$ is Hausdorff, then the compactness of the RHS implies its closure, hence: $$\overline{\phi(\mathcal{F})}\subseteq\phi(\overline{\mathcal{F}})$$That is, $\overline{\phi(\mathcal{F})}$ is a closed subset of a compact set, so it itself compact (this is true in any space).

That makes $\phi(\mathcal{F})$ precompact, as desired.

In fact, from this, it would follow that $\phi(\mathcal{F})=\phi(\overline{\mathcal{F}})$, so $\phi(\mathcal{F})$ is itself closed and compact.