A common exercise in the topology of the real line is to show that every family of pairwise disjoint (nonempty) open subsets of $\mathbb{R}$ is countable.

  • One quick way to show this is to note that every nonempty open subset must contain a rational number, and that no two open subsets in such a family can contain the same rational number, and so we get an injection from the family into $\mathbb{Q}$, meaning that the cardinality of the family cannot exceed $| \mathbb{Q} | = \aleph_0$.

In this proof there is nothing special about $\mathbb{R}$. We only use the fact that it is separable.

Another way of looking at "pairwise disjoint" families of nonempty subsets of some set $S$ is to note that each $x \in X$ is contained is at most one member of the family. Call such families "point-1" families.

By a "point-countable" family of subsets of $X$ we will mean a family $\{ A_i : i \in I \}$ of nonempty subsets of $X$ such that for each $x \in X$ the family $\{ i \in I : x \in A_i \}$ is countable.

A modification of the argument for the countability of point-1 families of open subsets of $\mathbb{R}$ shows that point-countable families of open subsets of $\mathbb{R}$ are also countable.

  • If $\{ U_i : i \in I \}$ is a point-countable family of (nonempty) open subsets of $\mathbb{R}$, for each $i \in I$ choose some $q_i \in U_i \cap \mathbb{Q}$. If $I$ were uncountable, then there would be a rational number $q$ such that $q_i = q$ for uncountably many $i \in I$. But then the family is not point-countable!

Clearly this argument can be trivially modified to show that in separable spaces point-countable families of open sets are countable.

The question

Are these two topological properties (the countability of all point-1 families of open sets, and the countability of all point-countable families of open sets) different? Or, to be more precise, is there a topological space $X$ such that

  1. all point-1 families of open sets in $X$ are countable, but
  2. there is an uncountable point-countable family of open sets in $X$?

By the above, such a space cannot be separable.


The space $X$ has countable cellularity $c(X)$ but uncontable Šhanin number $\check s(X)$. See below quotations from “General Topology” by Ryszard Engelking (2nd ed., Heldermann, Berlin, 1989).

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So as $X$ we can take the $\Sigma$-product of $\frak c$ many copies of a two-point discrete space $\{0,1\}$. Indeed, let $\hat X=\{0,1\}^\frak c$ and $X$ be the set of all elements $(x_\alpha)\in\hat X$ which have at most countably many non-zero coordinates. Since $X$ is dense in $\hat X$, $c(X)=c(\hat X)=\omega$ by Theorem 2.3.17. From the other hand, for each $\beta<\frak c$ let $U_\beta$ be the set of all elements $(x_\alpha)\in X$ whose $\beta$-th coordinate is $1$. Then the family $\{U_\beta:\beta<\frak c\}$ of non-empty open subsets of the space $X$ has cardinality $\frak c$, but the intersection of each its uncountable subfamily is empty.


Your point countable property is usually formulated using the notion of calibre: A space $X$ has calibre $\aleph_1$ whenever we have a family $\mathcal{U}$ of non-empty open sets of size $\aleph_1$, there is an uncountable subfamily with a common point. A moment's thought reveals that this is reformulation of "every point-countable subfamily is at most countable". It is classical that any product of spaces with calibre $\aleph_1$ again has that property, which follows from the so-called $\Delta$-system lemma in set theory.

See this blog post for more info. The square of a Suslin line would be a very nice example of a non ccc space of calibre $\aleph_1$, e.g.


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