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|. $

  • $\begingroup$ 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 $\emptyset$. $\endgroup$ Sep 15, 2020 at 22:15
  • $\begingroup$ 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. $\endgroup$ Sep 15, 2020 at 23:06
  • $\begingroup$ 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. $\endgroup$
    – Ulli
    Sep 16, 2020 at 7:48
  • $\begingroup$ 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. $\endgroup$
    – Ulli
    Sep 16, 2020 at 9:30
  • $\begingroup$ @Ulli it’s in the first book by Juhasz, it seems to be omitted in the second one. $\endgroup$ Sep 16, 2020 at 21:43

1 Answer 1


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.


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