# 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). –  Arthur Fischer 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

## 1 Answer

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!

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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
I don't see a proof?? You just claim it. –  Henno Brandsma Feb 20 '13 at 13:07
You are right. I am a little "of course". –  Paul Feb 20 '13 at 13:31