It is known that under $MA+ \neg CH$, every Corson compact space with the countable chain condition (ccc) is merizable. It is also known that, under $CH$, there exist nonmetrizable Corson compact spaces with ccc. Is it consistent with $ZFC+ \neg CH$ the existence of a nonmetrizable Corson compact space with ccc?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
Yes, take any model of $\neg CH$ where a Suslin Line exists (for example, start with a ground model where $CH$ does not hold and use Tennenbaum's original forcing for adding a Suslin Tree).
We can assume that our Suslin Line is compact. Indeed, let $Y$ be its Dedekind completion. Then $Y$ is a compact ccc linearly ordered topological space. We claim that $Y$ is not separable. Note that every ccc linearly ordered space is hereditarily Lindelof, so in particular, $Y$ has points $G_\delta$. Since $Y$ is compact, this implies that $Y$ is first-countable. If $Y$ were separable then, we could fix a countable dense subset $D$ of $Y$ and exploit first-countability to take, for every $x \in D$, a sequence $S_x$ inside our original Suslin Line, which converges to $x$. But then we would get that $\bigcup \{S_x: x \in D \}$ is a countable dense subset of our original Suslin Line, which is a contradiction. So $Y$ is a compact (first-countable) Suslin Line
Since $Y$ is a compact first-countable space, by a theorem of Shapirovskii (see Juhasz's Cardinal Functions book), we can find a cardinal $\kappa$ and an irreducible continuous mapping $f$:
$$f: Y \to \Sigma(\mathbb{R}^\kappa):=\{x \in \mathbb{R}^\kappa: |x^{-1}(1)| \leq \aleph_0 \}$$
The image $f(Y)$ is clearly a ccc Corson compact, and since irreducible maps preserve density, $f(Y)$ is not separable. So $f(Y)$ is a ccc non-metrizable Corson compact.
