# Product of Mrówka space and one point compactification discrete space.

I was reading an article and I have some troubles to understand it. First, the required definition to understand the problem:

Let $$\mathcal{U}\subseteq \{A\subseteq\omega: |A|=\aleph_0 \}$$. We say that $$\mathcal{U}$$ is an almost disjoint family if for all $$A,B\in\mathcal{U}$$ such that $$A\neq B$$ we have that $$|A\cap B|<\aleph_0$$

The proof that I was reading is the next:

The key part of the proof is the fact that $$A$$ is a closed subset of $$X\times Y$$. But I can't see that $$A$$ is closed only by the construction of the topology of $$X\times Y$$. In fact, I think that we need a lot of cases to prove that fact because if we take $$(a,b)\in (X\times Y)\setminus A$$ then

1. $$b=d^{*}$$.
2. $$a=r_\alpha$$ and $$b=d_\beta$$ with $$\alpha\neq\beta$$. Here probably we have two subcases because $$\alpha<\beta$$ or $$\beta<\alpha$$.
3. $$a\in\omega$$ and $$b=d_\alpha$$ for some $$\alpha<\mathfrak{c}$$
4. $$a\in\omega$$ and $$b=d^*$$.

Are they all cases? Or am I forgetting some? I don't know if my thoughts are correct. Can you help me to complete the proof? I really appreciate any help you can provide me.

• Note that lemma 2.1 is false, see my answer. – Henno Brandsma Apr 19 at 7:43
• And thus the main result's example fails: the author has not given an example of a quasi-Lindelöf space $X$ (as $X$ is weakly Lindelöf, not quasi-Lindelöf). Of course $X \times Y$ is not quasi-Lindelöf as it contains $X$ as a closed subset. So there is no example. – Henno Brandsma Apr 19 at 9:06
• It's still (IMHO) interesting whether there is a quasi-Lindelöf space $X$ whose product with a compact space is no longer quasi-Lindelöf. But this attempt at a counterexample fails. – Henno Brandsma Apr 19 at 10:47
• It is even more interesting MSE can help to spot flawed papers. Cheers! – YuiTo Cheng Apr 19 at 10:54

Suppose that $$(x,y) \notin A$$. If $$y \neq d^\ast$$, then $$y=d_\alpha$$ for some $$\alpha < \mathfrak{c}$$ while then $$x \neq r_\alpha$$. But then taking the neighbourhoods $$U_x=\{r_\beta\} \cup r_\beta$$ (if $$x=r_\beta$$ for some $$\beta\neq \alpha$$) or $$U_x =\{x\}$$ (if $$x \in \omega$$) and $$V_y=\{d_\alpha\}$$ we have that $$U_x \times V_y$$ also misses $$A$$, as $$V_y$$ only contains $$y$$ so the only way it could intersect $$A$$ is when $$r_\alpha \in U_x$$ which is clearly not the case by construction.

So the case that $$y \neq d^\ast$$ has been covered. So suppose $$y=d^\ast$$ and we need a neighbourhood of $$(x,y)$$ that misses $$A$$. If $$x \in \omega$$ take $$\{x\} \times Y$$ which clearly works as $$A$$ has no points with first coordinate in $$\omega$$, and if $$x=r_\beta$$ for some $$\beta$$, then it's easy to see that $$(\{r_\beta\}\cup r_\beta) \times (Y \setminus \{ d_\beta \})$$ is basic open and misses $$A$$ (as the neighbourhood of $$r_\beta$$ contains no other $$r_\alpha$$ by definition, just $$r_\beta$$ and some isolated points in $$\omega$$).

Just a word of warning:

Lemma 2.1 is false, and $$X$$ itself is a counterexample: $$\mathcal{R}$$ is closed and uncountable and discrete so not weakly Lindelöf. So not every closed subset of $$X$$ is weakly Lindelöf, so $$X$$ is separable ($$\omega$$ is dense) but not quasi-Lindelöf.

True is: a separable space is weakly Lindelöf. This is rather trivial to prove. The reference is not from a "proper" journal, so be warned...

Theorem 3.37 in the referenced paper [5] is also false (ccc implies quasi-Lindelöf) by the same counterexample. Don't believe everything in every random journal... I think this is the source of this paper's lemma 2.1 as I did not find a theorem on separable spaces (but of course separable implies ccc).

Also: 3.36 in [5] (every weakly Lindelöf normal space is quasi-Lindelöf) has a ZFC counterexample: $$C_p(X)$$ where $$X$$ is the one-point Lindelöfication of an uncountable discrete space (which can be seen to be the same as the $$\Sigma$$-product of uncountably copies of $$\mathbb{R}$$). Then $$C_p(X)$$ is (collectionwise) normal and is ccc; (it even has a dense Lindelöf subspace,) hence is weakly Lindelöf, but has a closed subspace homeomorphic to $$\omega_1$$ and hence is not quasi-Lindelöf (the cover by intial segments witnesses that $$\omega_1$$ is not weakly Lindelöf). For more details see this blog post, an excellent source for examples as these...

• +1 for the warning! – YuiTo Cheng Apr 19 at 7:22
• @YuiToCheng Your proof of closedness was simpler (but I wanted to show the claim by the most elementary means),+1 for noting it anyway. These papers the OP quotes (or indirectly quotes) are full of errors... – Henno Brandsma Apr 19 at 7:42
• Just out of interest, are you Dan Ma? – YuiTo Cheng Apr 19 at 12:03
• @YuiToCheng no but he writes blog posts that are in my sphere of interest. – Henno Brandsma Apr 19 at 12:06

Let $$f:\mathcal R\longrightarrow Y$$ be $$f(r_\alpha)=d_\alpha$$. $$f$$ is clearly continuous because $$\mathcal R$$, as a subspace of $$X$$, is discrete. Now, we claim that the graph of $$f=\{\langle r_\alpha,d_\alpha\rangle \mid \alpha<\mathfrak{c}\}$$ is closed in $$\mathcal R\times Y$$, which actully follows from this elementary theorem f is Continuous if and only if its Graph is Closed in 𝑋×𝑌 (here we only need the Hausdorffness of $$Y$$). As $$\mathcal R$$ is closed in $$X$$, $$\mathcal R \times Y$$ is a closed subset of $$X\times Y$$. Hence $$\{\langle r_\alpha,d_\alpha\rangle \mid \alpha<\mathfrak{c}\}$$ is closed in $$X\times Y$$.