Does there exist a topological space which is not separable but has the Souslin property?

Question

Does there exist a topological space which is not separable but possess the Souslin property (i.e. there is no uncountable family of pairwise disjoint nonempty open sets)? I guess it should be something really large, e.g. $[0,1]^\mathbb{R}$ or even $[0,1]^\mathbb{2^\mathfrak{c}}$ Thanks in advance.

Answer 1

Yes, the product is a nice and standard example:

$[0,1]^I$ is ccc (i.e. has the Souslin property) for any $I$. This is a corollary of the theorem that a product $\prod_i X_i$ is ccc iff all finite subproducts $\prod_{i \in F} X_i$ are, where $F \subseteq I$ is finite). This is a consequence of the $\Delta$-system lemma. And as $[0,1]$ is separable, and so are its finite powers, we see that $[0,1]^I$ is ccc.

And $[0,1]^I$ is separable iff $|I| \le \mathfrak{c}$, so $[0,1]^{\mathfrak{c}^+}$ is an example.

Another example is $C_p(X)$ (the space of real-valued continuous functions on a Tychonoff space $X$, in the topology induced by the product $\mathbb{R}^X$), with $X$ a compact non-metrisable Hausdorff space, (like the Double arrow space or the one-point comPactification of an uncountable discrete space etc.), because all spaces of the form $C_p(X)$ are ccc (being dense in $\mathbb{R}^X$) and $d(C_p(X)) = w(X)$ for $X$ compact Hausdorff). This is another, but related, way to construct such spaces.

The classic is of course a Souslin line (the reason we call ccc the Souslin property, in a way), which exists in some but not all models of ZFC: a linearly ordered space (order-dense, unbounded) which in the order topology is ccc but is not separable.

The most boring example: $\mathbb{R}$ in the co-countable topology. This is not separable as all countable sets are closed and thus not dense and ccc because all non-empty open sets intersect.

Answer 2

Yes, $[0, 1]^{2^\mathfrak{c}}$ works, but $[0, 1]^ℝ$ is not enough. The point is that any product of separable spaces is ccc, because a product is ccc if and only if its every finite subproduct is ccc. At the same time the cardinality of a Hausdorff space $X$ is upper bounded by $2^{2^{d(X)}}$.

Also, the Hewitt–Marczewski–Pondiczery theorem says that a product of $≤2^κ$ many spaces of density $≤ κ$ still has density $≤ κ$. So $[0, 1]^ℝ$ is still separable.

In fact, it can be proved that $d(∏_{i ∈ I} X_i) = \log|I| ∨ \bigvee_{i ∈ I} d(X_i)$ if the spaces are Hausdorff and nondegenerate (and the vees denote suprema).