In the thesis "[Topology of Function Spaces](http://d-scholarship.pitt.edu/7814/)" 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](https://mathoverflow.net/questions/257686/lindel%C3%B6f-property-for-open-covers-for-compact-sets/257757#257757)), 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).
> 
> 
> 



Why is $C_k(\omega_1)$ Lindelöf?
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 $g\in B\_{\alpha\_f}[f;3\varepsilon\_f)\subset U\_f$.


