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If we start with $V\models\lozenge$, it is not hard to force the failure of diamond. You can blow up the continuum, or destroy all the Suslin trees. You can blow up the continuum of $\aleph_1$, and then collapse $\aleph_1$ to be countable.

There are many ways of doing that, but all of them (that I could think of, with the help of a few people over the day) include one of the two:

  1. Blowing up the continuum,
  2. Collapsing cardinals.

Is it consistent that $V\models\lozenge$, and $r$ is a $V$-generic real such that $V[r]\models\lnot\lozenge+\sf CH$ and no cardinals were collapsed between $V$ and $V[r]$?

If the answer is positive, can we strengthen the preservation of $\sf CH$ by requiring also that the continuum function remains the same (so no blowing up power sets of larger cardinals somehow)?

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    $\begingroup$ Possibly worth mentioning: if $\mathbb{P}$ is ccc and $|\mathbb{P}| \leq \aleph_1$, then forcing with $\mathbb{P}$ preserves $\diamondsuit$ and CH. This is Exercise IV.7.58 in the newer set theory book by Kunen. $\endgroup$
    – Will Brian
    Commented May 26, 2015 at 17:12
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    $\begingroup$ Jensen's forcing for $SH+CH$ forces the failure of diamond without collapsing cardinals or blowing up the continuum, but its generic is certainly not a real. $\endgroup$
    – Yair Hayut
    Commented May 26, 2015 at 17:51
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    $\begingroup$ I've upvoted in order to partially offset the inevitable backlash from environmentalists who react negatively to "blow up the continuum and destroy all the Suslin trees." $\endgroup$
    – Vidit Nanda
    Commented May 26, 2015 at 19:47
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    $\begingroup$ Do you know the answer if we weaken $r \subseteq \omega$ to $r \subseteq \omega_1$? $\endgroup$
    – Monroe Eskew
    Commented May 27, 2015 at 2:56
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    $\begingroup$ The answer to your question is no, if $V=L,$ as then for any real $r, L[r]\models \Diamond+CH.$ $\endgroup$
    – Mohammad Golshani
    Commented May 27, 2015 at 4:08

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The answer is yes, assuming the existence of $\aleph_2$-many measurable cardinals. To see this, assume $GCH+\Diamond$ holds and $S$ is a discrete set of measurable cardinals of size $\aleph_2.$

Step 1. Force with Prikry product forcing $P_S$ to change the cofinality of each element of $S$ into $\omega.$

Note that the extension is of the form $V[(x_\alpha: \alpha\in S)]$, where each $x_\alpha$ is an $\omega-$sequence cofinal in $\alpha.$

Step 2. Force with Jensen's coding theorem, to code everything into a real $r$, so that we have $V[(x_\alpha: \alpha\in S)][r]=V[r]$ (we can do this using a set forcing construction, and assuming that the ground model is a core model).

Step 3. Force over $V[r]$, by a cardinal and $GCH$ preserving forcing iteration to force $\neg \Diamond$.

Note that the generic can be seen as a subset $X$ of $S.$ Now working in $V[r][X],$ define a new sequence $(y_\alpha: \alpha\in S),$ so that $y_\alpha=x_\alpha,$ if $\alpha\in X$ and $y_\alpha=x_\alpha\setminus\{ min(x_\alpha) \}$ if $\alpha\notin X.$

Then let $V_1=V[(y_\alpha: \alpha\in S)]$ and $V_2=V_1[r].$ Note that:

1) $V_1\models GCH+\Diamond,$

2) $V_2=V[(y_\alpha: \alpha\in S)][r]=V[r][X]\models GCH+\neg\Diamond,$

3) $V_2=V_1[r]$, for some real $r$.

I may mention that the above method can be used to prove the consistency of many statements using adding a single real.

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  • $\begingroup$ But is this $r$ generic over $V_1$? It seems class generic? (Although probably coding just a portion of the universe, not everything, would suffice, no?) $\endgroup$
    – Asaf Karagila ♦
    Commented May 27, 2015 at 12:55
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    $\begingroup$ The coding part can be done using a set forcing, as we are coding a set of ordinals. For example we can start with the ground model being a core model $K$, and we will have $K[(x_\alpha: \alpha<\omega_2)][r]=K[r].$ $\endgroup$
    – Mohammad Golshani
    Commented May 27, 2015 at 13:38
  • $\begingroup$ I think that I understand this construction, which is quite clever. You might want to change/remove the remark that $V[r]=L[r]$, since we don't want $r$ to be class-generic and it might not be the case that $r$ encodes the entire universe anymore. I'll sleep on it, and see if I have additional questions tomorrow morning. $\endgroup$
    – Asaf Karagila ♦
    Commented May 27, 2015 at 22:37
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    $\begingroup$ Nice argument. Can we justify the use of measurables here? $\endgroup$
    – Ashutosh
    Commented May 28, 2015 at 12:40

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