# Some contradiction in an open problem

Recently, I'm considerring an open problem from the paper by Ofelia T.Alas and others':On the extent of star countable spaces. It easily can be downloaded by google. The open Problem is this:

Suppose that $X$ is a (strongly) monotonically monolithic star countable space. Must $X$ be Lindelof?

Somebody has proved it is true that if $X$ is a strongly monotonically monolithic star countable space, then $X$ must be Lindelof. (see the paper: A note on the extent of two subclasses of star countable spaces by the author Zuoming Yu).

In the morning, I met acrossly an example See Dan Ma's discussion non-normal product space: there exist a separable metric space $X$ and a Lindelöf space $Y$, such that $X \times Y$ is not a Lindelöf space.

On one hand because every point-countable base space is strongly monotonically monolithic, according to Tkachuk, "monolithic spaces and D-spaces revisited", Proposition 2.5, then $X$ of course is strongly monotonically monolithic. And $Y$ is also have point-countable base, so $Y$ is also strongly monotonically monolithic. Moreover, $X\times Y$ is strongly monotonically monolithic (see the same paper of Tkachuk: theorem 2.10).

On the other hand, The space $X$ and the space $Y$ are two star countable, clearly, their product is also star countable.

Now the contradiction comes, such space $X \times Y$ is not normal, and hence is not Lindelof, which contradicts with the result of Zuoming Yu.

I don't know what's wrong. Could somebody help? Thanks for any help.

Added: The product of 2 star-countable spaces is star-countable:

Suppose $X$ and $Y$ are both star countable. Given any open cover of basic nbhd $U\times V=\{u\times v: u \in U; v\in V\}$ of $X\times Y$. Clearly, $U$ is an open cover of $X$ and $V$ is an open cover of $Y$, therefore, there exist countable subsets $C\subset X$ and $D\subset Y$ which satisfies that $St(C,U)=X$ and $St(D,U)=Y$. Now we can get an countable subset: $C\times D$ of $X\times Y$ and easily see that $St(C\times D, U\times V)=X\times Y$.

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Matveev's A Survey on Star Covering Properties appears to give some references to products of star-countable (what he calls star-Lindelof) spaces which are not star-countable (Examples 19, 20, and 21 on p.80). – arjafi Feb 20 '13 at 13:20
@ArthurFischer: Thanks for the information! – Paul Feb 20 '13 at 13:31
Not all covers are of the form you describe. In general we can reduce to covers of the form $\{U_i \times V_i : i \in I \}$, and the $U_i$ firm a cover of $X$, the $V_i$ of $Y$, but the not every combination $U_i \times V_j$ is in that cover, but this is what you suggest with $U \times V = \{ u \times v: u \in U, v \in V \}$. – Henno Brandsma Feb 20 '13 at 17:34

Do you have a reference or proof for "the product of 2 star-countable spaces is star-countable"? I don't see that. And your $X$ and $Y$ (the Michael line but based on a Bernstein set.. as $Y$) could be a counterexample.
As to your (added) proof: pick $(x,y) \in X \times Y$. We get independent $U$ and $V$ for the different coordinates, but we can not yet say that the product of these is in the originla cover of product sets!
However it is Lindelof, and hence it must be star countable. $Y$ is some different from the Michael line. Michael line is not Lindelof. – Paul Feb 20 '13 at 12:58