I am looking for a comprehensive list of known forcing notions which add a minimal real into the ground model. I know some of them like the Sacks forcing, or the Judah-Shelah's example of a c.c.c. forcing which adds a minimal real (and a few more).

Question. What are other known (or less known) examples of forcing notions which add a minimal generic real?

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    $\begingroup$ A close relative of Sacks reals are Jensen reals, which in addition to being minimal, are $\Delta^1_3$-definable. They were introduced in Jensen' paper: Definable sets of minimal degree. Mathematical logic and foundations of set theory (Proc. Internat. Colloq., Jerusalem, 1968), pp. 122–128. North-Holland, Amsterdam, 1970. $\endgroup$ – Ali Enayat Oct 12 at 18:44

Splitting forcing is a little-known forcing that adds a splitting real and creates a minimal extension. It consists of splitting trees ordered by inclusion. A splitting tree is a perfect tree $T \subseteq 2^{<\omega}$ where for every $s \in T$ there is $m \in \omega$ so that for all $n \geq m$, $i \in 2$, there is $t \in T$, $s \subseteq t$ with $t(n)= i$.

In my recent preprint (arxiv link), I show that splitting forcing adds a minimal real (Corollary 4.20) and provide part of an argument for the minimality of the forcing extension (Corrolary 3.21).

Other than that, splitting forcing is $\omega^\omega$-bounding but as far as I know it is unknown whether it has the Sacks-property (see also Question 6.4 in On splitting trees by Laguzzi, Mildenberger and Stuber-Rousselle).

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  • $\begingroup$ Thanks a lot for the answer. $\endgroup$ – Mohammad Golshani Oct 18 at 5:43

Prikry-Silver forcing adds a minimal real: that is, if $g$ is Prikry-Silver generic over $V$ and $h$ is any real in $V[g] \setminus V$, then $V[g] = V[h]$.

Interestingly, while Prikry-Silver forcing is minimal for reals, it is not fully minimal (like Sacks forcing). If $g$ is Prikry-Silver generic over $V$, then there is a set $F \in V[g] \setminus V$ such that $V[F] \neq V[g]$. (Specifically, this is true when $F$ is the filter generated by all complements of domains of conditions in the generic; or, equivalently, if $g$ is the Prikry-Silver generic real, then $F$ is the filter generated by $\{ A \in V \cap \mathcal P(\omega) : g \restriction A \notin V \}$. Then $F$ is a new non-principal ultrafilter on $\omega$, and $V[F]$ contains no new reals.)

Edit: The minimality-for-reals of Prikry-Silver forcing can be seen as a special case of the following theorem of Grigorieff:

If $\mathcal U$ is a selective ultrafilter, then (what's now called) the Grigorieff forcing on $\mathcal U$ adds a minimal real. The conditions in this forcing are the partial functions $A \rightarrow 2$ where $A \notin \mathcal U$, and the ordering is the natural one (extension).

The reason this implies Prikry-Silver forcing adds a minimal real (but is not fully minimal) is that Prikry-Silver forcing can be factored as a $2$-step iteration: first force with $\mathcal P(\omega)/\mathrm{fin}$ to add a selective ultrafilter, and then force with the Grigorieff forcing on that ultrafilter. The first step adds no new reals, and the second step adds a minimal real by the theorem quoted above.

(All of this stuff can be found in this paper of Grigorieff. This old MO thread is also relevant.)

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  • $\begingroup$ Can you see Prikry–Silver as an iteration of adding a new ultrafilter, then adding a real splitting it somehow or something? $\endgroup$ – Asaf Karagila Oct 12 at 14:49
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    $\begingroup$ @AsafKaragila: Yes, I think. I'll need to check to be sure, but I think that adding a Prikry-Silver real is equivalent to: (1) forcing with $\mathcal P(\omega) / \mathrm{fin}$ to add a new ultrafilter, and then (2) adding a new real using the Grigorieff forcing associated to the new ultrafilter. $\endgroup$ – Will Brian Oct 12 at 15:00
  • $\begingroup$ @WillBrian Thanks Will for the answer. May you also give a reference. $\endgroup$ – Mohammad Golshani Oct 18 at 5:42
  • $\begingroup$ @MohammadGolshani: See this paper of Grigorieff -- citeseerx.ist.psu.edu/viewdoc/…. $\endgroup$ – Will Brian Oct 20 at 14:06

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