large ccc forcing that preserves CH

Can you name a ccc forcing with the following properties?

1) Atomless and separative

2) The least size of a dense set is large, say at least $\aleph_3$, hopefully as big as you like.

3) Existence is consistent with CH.

4) Preserves CH.

If you can think of any, please list as many as you can.

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Thanks to Michael Blackmon for this idea:

If P has the ccc and preserves CH, then there is a contiuum sized regular suborder R that adds all the reals P will add. The factor forcing P/R is ccc and $\omega$-distributive, a Suslin algebra. A theorem (of Jech?) says that any Suslin algebra has size at most $2^{\omega_1}$. So there is a bound on the size of P.

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Very nice idea. I guess you mean that it is $(\omega,2)$-distributive, rather than $(\omega,\infty)$-distributive, since not adding reals is only equivalent to $(\omega,2)$-distributivity, not $(\omega,\infty)$-distributivity. Is this enough for the Suslin algebra theorem you mention? – Joel David Hamkins Oct 5 '12 at 21:56
This is great! It would be nice to have it fleshed out. (Here or elsewhere, but it would be nice if Michael made a comeback on MO...) – François G. Dorais Oct 6 '12 at 2:25
Joel, since we have the ccc here, $(\omega,\infty)$-distributivity is equivalent to $(\omega,\omega)$-distributivity. Easy to see using the definition in terms of common refinements of sets of antichains. – Monroe Eskew Oct 6 '12 at 20:10

Here is a way to produce examples with arbitrarily large densities.

Let $\mathbb{P}$ be any c.c.c. forcing that preserves CH, such as the forcing to add a Cohen real, and let $\mathbb{Q}=\kappa^\ast$ be the decreasing linear order of length $\kappa$, a large regular cardinal. Consider the product partial order $\mathbb{P}\times\mathbb{Q}$ as a notion of forcing. This is still c.c.c., since all conditions in the second coordinate are compatible as it is linear. But every dense set must also be dense in the second coordinate and therefore have size at least $\kappa$. Since the second coordinate is trivial as a forcing notion, the product $\mathbb{P}\times\mathbb{Q}$ is forcing equivalent to $\mathbb{P}$, which preserves CH, and so $\mathbb{P}\times\mathbb{Q}$ has all your desired properties.

This kind of example, however, may lead you to modify the question, since it is not separative.

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Good point, I will modify it. – Monroe Eskew Sep 20 '12 at 4:54