I have been reading about cardinal invariants and I have a question about the independence number $\mathfrak{i}$. In Blass's paper (Combinatorial Characteristics of the Continumm) it is mention that in Sacks model, Eisworth and Shelah proved that $\mathfrak{i}= \aleph\_1$, unfortunately it mentions that the result is unpublished, I want to know if someone knows any source to found it or maybe can give me an idea of how it works.
Do they construct a maximal independent family in the ground model of size $\aleph\_1$ that remains maximal after the iteration?, It is constructed used names as in the case $\mathfrak{a}=\aleph\_1$?


Thanks for your help! 



The independence number
This takes me back to Jerusalem in the late 90s. I've actually been asked about this a couple of times this year after 15 years of hearing nothing about it.


The proof consisted of looking at the maximal independent family constructed by Shelah in the course of his proof of Con($\mathfrak{i}<\mathfrak{u}$). The m.i.f. he builds there (assuming CH) is designed to survive the forcing he uses to push up $\mathfrak{u}$.


Sacks forcing is much gentler than the forcing Shelah uses to obtain $\frak{i}<\frak{u}$, and my recollection is that the proof that the special m.i.f. survives iterated Sacks forcing was "just" Shelah's proof used to obtain $\mathfrak{i}<\mathfrak{u}$, with the details being much easier. I also recall that Shelah's proof was certainly a proof by Shelah...


Think about what is needed: keeping the m.i.f. alive during the iteration amounts to preserving the unsplittability of a naturally related collection of sets (see the proof that $\mathfrak{r}\leq\mathfrak{i}$), and this is close to preserving ultrafilters. Shelah's maximal independent family is constructed so that the relevant unsplittable family shares many properties with Ramsey ultrafilters, and that is the key to making sure its unsplittability is preserved.


Sorry I can't be more specific --- it was a long time ago, and didn't really seem to be of interest at the time given that all the ideas were already in Shelah's paper! (I haven't had success finding old notes, but reconstructing this is on my plate for the summer.)




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Update 2-17-17: My graduate student Michael Perron noted that there is a fairly easy way to see that $\mathfrak{i}$ is $\aleph\_1$ in the Sacks model.


If $I$ is an independent family, let $env(I)$ (the *envelope of $I$*) denote the collection of all finite intersections of elements of $I$ and complements of elements of $I$ (throwing out anything that comes up empty, of course).


$I$ is *selective* if for every $f:\omega\rightarrow\omega$ there is an $A\in env(I)$ such that $f\upharpoonright A$ is either constant or one-to-one.


A selective independent family if maximal (given $B\subseteq\omega$, let $f$ be the characteristic function of $B$ and look at what happens).


$I$ is *everywhere selective* if for any $f:\omega\rightarrow\omega$ and $A\in env(I)$ there is a $B\subseteq A$ in $env(I)$ such that $f\upharpoonright B$ is either constant or one-to-one.


$I$ is everywhere selective if and only if the ideal consisting of those sets $X\subseteq \omega$ such that for any $A\in env(I)$ there is a $B\subseteq A$ in $env(I)$ with $B\cap X=\emptyset$ is a selective ideal in the sense considered in the original Baumgartner-Laver paper that considers iterated Sacks forcing. They prove that selective ideals remain selective in the Sacks model (iterating Sacks forcing $\omega\_2$ times).


Now CH can be used to build an everywhere selective independent family $I$, and so we can get the result.


