# Is an $\omega_1$-Lindelöf space with a $G_\delta$- diagonal a Lindelöf space?

Suppose $X$ is an $\omega_1$-Lindelöf space with a $G_\delta$-diagonal. I want to show it is a Lindelöf space by the method of Chaber from 1976, by which he proved a countably compact space with a $G_\delta$-diagonal must be compact. I found it is also very useful to prove that $\omega_1$-Lindelöf spaces with a $G_\delta$-diagonal are Lindelöf. I believe that this is very close to my desired result. However, there is a stage I can't pass. Could some one give me some suggestions? I will write down the proof and I will highlight what is troubling me.

Prove: Let $X$ be an $\omega_1$-Lindelöf space and $\mathcal{G_n}$ be a $G_\delta$-diagonal sequence for $X$ such that $\mathcal{G}_{n+1}$ refines $\mathcal{G}_{n}$. Suppose that $X$ is not Lindelöf. Then there exists an open cover $\mathcal{U}$ such that $\mathcal{U}$ contains no $\omega_1$-sized subcover. Pick $x_0 \in X$. Observe that there exists $n(0) \in \omega$ such that $\mathcal{U}$ contains no $\omega_1$-sized subcover of $X \setminus \mathrm{St}(x_0, \mathcal{G_{n(0)}})$. Otherwise, for each $n$, we could find a $\omega_1$-sized $\mathcal{U_n}$ of $\mathcal{U}$ covering $X \setminus \mathrm{St}(x_0, \mathcal{G_n})$, and then $\bigcup_n\mathcal{U_n}$ would be an $\omega_1$-sized cover of $X\setminus \{x_0\}$, hence $\mathcal{U}$ would contain an $\omega_1$-sized cover of X. Contradiction!

Now we inductively choose, for each $\alpha \in \omega$, a point $x_\alpha \in X$ and $n(\alpha) \in \omega$ such that

$(i)$ $x_\alpha \in X\setminus \bigcup_{\beta<\alpha}\mathrm{St}(x_\beta, \mathcal{G}_{n(\beta)})$.

$(ii)$ $\mathcal{U}$ does not contain a $\omega_1$-sized cover of $X\setminus \bigcup_{\beta \leq \alpha}\mathrm{St}(x_\beta, \mathcal{G}_{n(\beta)})$.

Now I hope this can be done for $\omega$. If it can work on $\omega$, then for some $n\in \omega$ we can get uncountable subset $A$ of $\omega_1$ such that $n(\beta)=n$ for each $\beta \in A$. Then $\{x_\beta： \beta \in A\}$ is an uncountable closed discrete subset of $X$, for each $G\in \mathcal{G_n}$ meets at most one element of this set. This contradicts the countable extent (from $\omega_1$-Lindelöf).

However, I failed to make the above process work on $\omega$. Could someone help me save it (maybe by adding some condition)?

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To save duplicated effort, it would be good manners to link to the version of this (or the related question) that has been asked on MathOverflow –  user16299 Jan 7 '12 at 5:55