Compact scattered spaces $X$ with $w(X)<|X|$

Let $$X$$ be a topological space. For any set $$S$$, let us denote its cardinality by $$|S|$$. The weight of $$X$$ is defined by $$w(X) = \inf\{|\mathcal B|: \mathcal B\mbox{ is a basis for the topology of X}\}.$$ And let us recall that a topological space is called scattered if, and only if, every subspace $$A\subset X$$ has an isolated point (with respect to $$A$$).

My question is whether there exists a compact scattered space $$X$$ such that $$w(X) < |X|.$$

• I think you can show that $|X| \le w(X)$ for scattered spaces (under some mild separation axiom maybe). Follow the scattering of $X$ down to $\emptyset$. Sep 15, 2020 at 22:15
• I guess it can be showed by transfinite indcuction that the fundamental neighborhood sistem of any ordinal $\alpha$ must have at least the cardinality of $\alpha$, so $w(\alpha)=|\alpha|$ for any ordinal $\alpha$. Then, considering $(\alpha, n)$ the Cantor-Bernstein index of $X$, since there is always an homeomorphic embedding $\alpha\cdot n +1 \to X$, we conclude that $w(X) \geq |\alpha|$. In order to finish this argument I would need to show that $|\alpha|=|X|$, but I can't see how. Sep 15, 2020 at 23:06
• I think for any infinite scattered space (without any further separation requirements), it holds $hL(X) = |X|$, where $hL(X)$ is the hereditarily Lindelof degree of $X$. Of course, $hL(X) \le w(X)$. Probably this can be proven using a right separating well order on $X$. Might be that I.JUHÁSZ, CHAPTER 2 - Cardinal Functions II in Handbook of Set-Theoretic Topology 1984 contains a prove. Unfortunately, I don't have access to this book anymore.
– Ulli
Sep 16, 2020 at 7:48
• I assume I was wrong with the above citation. I checked this article here, but didn't find the result. But I'm pretty sure it already appeared somewhere in the literature.
– Ulli
Sep 16, 2020 at 9:30
• @Ulli it’s in the first book by Juhasz, it seems to be omitted in the second one. Sep 16, 2020 at 21:43

1 Answer

The book Cardinal Functions in Topology By I. Juhasz (I have the 1971 edition) proves that for scattered spaces $$X$$ we have $$h(X)=|X|$$ (this is Theorem 2.14 on p. 22) where $$h(X)$$ is the sup of all order types of right-separated subspaces (a subspace $$S$$ is right-separated iff it has a well-order $$<_S$$ such that all initial segment sets $$\{x \in S: x<_S s\}$$ are open in $$S$$). The proof comes from the scattering levels of $$X$$; picking one point from each gives us a right-separated subspace. One can show that $$h(X)$$ equals the hereditarily Lindelöf number $$hL(X)$$ of $$X$$ and $$hL(X) \le w(X)$$ is trivial.

So $$|X| \le w(X)$$ follows from those facts. So we cannot have $$w(X) <|X|$$, even without the compactness.