TITLE: Why is $C_k(\omega_1)$ Lindelöf?
QUESTION [6 upvotes]: In the thesis "Topology of Function Spaces" by Andrew Marsh (Galway, 2000) it is claimed (page 23, before Definition 30) that the space $C(\omega_1)$ is Lindelöf with respect to the compact open topology. For a proof the reader is refered to 
S.P. Gul'ko. On properties of $\sigma$-products. Soviet Math. Dokl. 18, 6 1977.
I searched a while on the net but I could not find another reference to this paper - it seems there is a paper by Gul'ko named "On properties of subsets of $\Sigma$-products" but I do not know wether that is the same paper or how to obtain access to the latter paper.
I wanted to ask Andrew Marsh but I could not find a way to address him (plus there seem to be more people of that name).
Furthermore, I could not find any other source for that claim.
So, my question is:


Can you give me a reference (or a proof if it fits on one page) for the fact that $C_k(\omega_1)$ is Lindelöf? It does not need to be the Gul'ko paper above but if you manage to find it and it really contains the proof, that is of course also very fine.
Bonus question (not really part of the question but it would be even nicer): Is it known if $C_k(\omega_1) $ is $k$-Lindelöf in the sense of my last question (see Lindelöf Property for Open Covers for Compact Sets), i.e. does every open $k$-cover admit a countable $k$-subcover? (a $k$-cover is a cover such that every compact subset is contained in one of the members of the cover).

REPLY [4 votes]: Here is (a bit lengthy and technical) proof of the Lindelof property of the function space $C_k(\omega_1)$. At first some notations.
For any function $f\in C_k(\omega_1)$ and a countable ordinal $\alpha$ let $\|f\|_\alpha=\sup_{x\in[0,\alpha]}|f(x)|$. Let also $\|f\|=\sup_{x\in\omega_1}|f(x)|$.
For every $f\in C_k(\omega_1)$, $\alpha\in\omega_1$ and $\varepsilon>0$ consider the open neighborhood $$B_\alpha[f;\varepsilon):=\{g\in C_k(\omega_1):\|g-f\|_\alpha<\varepsilon\}$$ of $f$ in the function space $C_k(\omega_1)$.
Given an open cover ${\mathcal U}$ of $C_k(\omega_1)$, for every $f\in C_k(\omega_1)$ let $$
\begin{aligned}
\varepsilon_f&:=\sup\{\varepsilon\in (0,1]:\exists \alpha\in\omega_1\;\exists U\in{\mathcal U}\;\;B_\alpha[f;4\varepsilon)\subset U\},\mbox{ and }\\
\alpha_f&:=\min\{\alpha\in\omega_1:\exists U\in{\mathcal U}\;\;B_\alpha[f,3\varepsilon_f)\subset U\}.
\end{aligned}
$$
Choose also a set $U_f\in{\mathcal U}$ such that $B_{\alpha_f}[f;3\varepsilon_f)\subset U_f$.
Claim. For any functions $f,g\in C_k(\omega_1)$ we get $4\varepsilon_f\ge 4\varepsilon_g-\|f-g\|$.
Proof. Assuming that $4\varepsilon_f<4\varepsilon_g-\|f-g\|$, we can choose $\delta>0$ such that  $\|f-g\|+4\varepsilon_f+\delta\le 4\varepsilon_g-\delta$ and find an ordinal $\alpha\in\omega_1$ such that $B_\alpha[g;4\varepsilon_g-\delta)\subset U$ for some $U\in{\mathcal U}$. Then $$B_\alpha[f;4\varepsilon_f+\delta)\subset B_\alpha[g;\|f-g\|+4\varepsilon_f+\delta)\subset B_\alpha[g;4\varepsilon_g-\delta)\subset U\in{\mathcal U},$$ which contradicts the definition of the number $\varepsilon_f$.  $\square$
For every ordinal $\alpha\in\omega_1$, identify the Banach space $C_k[0,\alpha]$ with the subspace $\{f\in C_k(\omega_1):f|[\alpha,\omega_1)\equiv const\}$ of $C_k(\omega_1)$, consisting of functions, which are constant on the interval $[\alpha,\omega_1)$.
For two ordinals $\alpha,\beta$ by $\alpha{\vee}\beta$ we denote their maximum $\max\{\alpha,\beta\}$.
We shall construct inductively a non-decreasing sequence of countable ordinals $(\alpha_n)_{n\in\omega}$ and a sequence $(F_n)_{n\in\omega}$ of countable subsets $F_n\subset C_k[0,\alpha_n]$ such that for every $n\in\omega$ the following conditions are satisfied:
$(1_n)$ $C_k[0,\alpha_n]\subset \bigcup_{f\in F_n}B_{\alpha_n\!{\vee}\alpha_f}[f;\varepsilon_f)$;
$(2_n)$ $\alpha_{n+1}=\sup\limits_{f\in F_n}(\alpha_n{\vee}\alpha_f)$.
We start the inductive construction letting $\alpha_0=0$. Assume that for some $n\in\omega$ an ordinal $\alpha_n$ has been constructed. For the open cover $\{B_{\alpha_n\!{\vee}\alpha_f}[f;\varepsilon_f):f\in C_k[0,\alpha_n]\}$ of the separable Banach space $C_k[0,\alpha_n]$, there exists a countable subset $F_n\subset C_k[0,\alpha_n]$ such that $C_k[0,\alpha_n]\subset \bigcup_{f\in F_n}B_{\alpha_n{\vee}\alpha_f}[f;\varepsilon_f)$. Letting $\alpha_{n+1}:=\sup_{f\in F_n}(\alpha_n{\vee}\alpha_f)$ we complete the inductive step.
After completing the inductive construction, consider the countable ordinal $\alpha_\omega=\sup_{n\in\omega}\alpha_n$ and the countable set $F:=\bigcup_{n\in\omega}F_n\subset C_k[0,\alpha_\omega]\subset C_k(\omega_1)$. We claim that $\{U_f:f\in F\}\subset {\mathcal U}$ is a countable subcover of $C_k(\omega_1)$. Given any function $g\in C_k(\omega_1)$, for every ordinal $n\le\omega$  consider the (unique) function $g_n\in C_k[0,\alpha_n]$ such that $g_n|[0,\alpha_n]=g|[0,\alpha_n]$.
By the continuity of the function $g$ at $\alpha_\omega$, there exists a number $n\in\omega$ such that $|g(x)-g(\alpha_\omega)|<\varepsilon_{g_\omega}$ for all $x\in[\alpha_n,\alpha_\omega]$. This implies that  $\|g_n-g_\omega\|<\varepsilon_{g_\omega}$ and $4\varepsilon_{g_n}\ge 4\varepsilon_{g_\omega}-\|g_n-g_\omega\|>3\varepsilon_{g_\omega}$ according to the Claim.
By the inductive condition $(1_n)$, for the function $g_n\in C[0,\alpha_n]$ there exists $f\in F_n$ such that  $g_n\in B_{\alpha_n\!{\vee}\alpha_f}[f;\varepsilon_f)$ and hence  $\|f-g_n\|_{\alpha_n\!{\vee}\alpha_f}<\varepsilon_f$. By the inductive condition $(2_n)$, $\alpha_f\le\alpha_{n+1}\le\alpha_\omega$. Since the functions $f$ and $g_n$ are constant on the interval $[\alpha_n,\omega_1)$, the inequality $\|f-g_n\|_{\alpha_n\!{\vee}\alpha_f}<\varepsilon_f$ implies $\|f-g_n\|=\|f-g_n\|_{\alpha_n\!{\vee}\alpha_f}$. Then Claim yields that  $4\varepsilon_{f}>4\varepsilon_{g_n}-\|f-g_n\|>3\varepsilon_{g_n}$ and hence $\varepsilon_{g_\omega}<\frac43\varepsilon_{g_n}<\frac43\frac43\varepsilon_{f}$.
We claim that $g\in B_{\alpha_f}[f;3\varepsilon_f)\subset U_f$.  Indeed, for every $x\in [0,\alpha_f]$ we get $$|g(x)-f(x)|\le |g(x)-g_n(x)|+|g_n(x)-f(x)|<\varepsilon_{g_\omega}+\varepsilon_f<\tfrac{16}9\varepsilon_{f}+\varepsilon_f<3\varepsilon_f,$$
 and hence $g\in B_{\alpha_f}[f;3\varepsilon_f)\subset U_f$.