Suppose that $V$ is a model of $\sf ZFC$, and fix some regular $\kappa$, say $\omega_1$ for practical purposes.

Let $\cal U$ be an ultrafilter on $\omega_1$ in $V$ which is non-principal and even uniform. Let $\Bbb P$ be a forcing such that:

  1. $\Bbb P$ does not add bounded subsets to $\kappa$, but its generic defines a canonical unbounded subset of $\kappa$, say $G$.
  2. $\Bbb P$ satisfies $\kappa^+$-c.c.
  3. $\Bbb P$ is homogeneous.

Of course that $\cal U$ is not an ultrafilter in $V[G]$ since it is no longer closed under supersets. But we can ask whether or not $\mathcal U\cup\{G\}$ is an ultrafilter base.

Is there a condition on $\cal U$ (or an additional condition on $\Bbb P$ or $G$) which guarantee that $\mathcal U\cup\{G\}$ is an ultrafilter base in $V[G]$?


Here is a necessary and sufficient condition:

Theorem. If $U$ is an ultrafilter in $V$, then the following are equivalent:

  1. $U\cup\{G\}$ is an ultrafilter base.
  2. $G$ is not disjoint from any element of $U$ and $U$ decides every subset of $G$ in $V[G]$, meaning that for every $A\subset G$ in $V[G]$ there is $B\in U$ with $B\cap G$ contained in $A$ or disjoint from $A$.

The proof is straightforward.

It follows that many of the usual forcing notions do not have this property. For example, if the forcing is adding a Cohen subset $G\subset\kappa$, then we definitely won't have condition 2. For example, if $A$ consists of every other element of $G$ and $B$ is any unbounded ground model set, then it is dense that $B\cap G$ is neither contained in nor disjoint from $A$.

  • $\begingroup$ Thanks! I think that might be helpful in understanding my situation a bit better! $\endgroup$
    – Asaf Karagila
    Nov 26 '15 at 16:00
  • 2
    $\begingroup$ Do you have an example of a forcing that has this property? $\endgroup$
    – Goldstern
    Feb 17 '16 at 17:13

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