# Does $Add(\kappa,1)^L$ ever collapse cardinals?

In general, we know that adding a subset to a regular cardinal $\kappa$ can collapse cardinals. If, for example, there is $\gamma < \kappa$ with $2^\gamma >\kappa$, then $Add(\kappa,1)$ will collapse $2^\gamma$ to $\kappa$, since every subset of $\gamma$ will appear as a block in the generic.

However, this argument requires that we use the poset defined in $V$. We can easily construct situations in which $2^\gamma > \kappa$ but forcing with $Add(\kappa,1)^L$ does not collapse cardinals, by the same techniques that allow us to achieve Easton's Theorem. If we force over $L$ with $Add(\gamma, \kappa^+) \times Add(\kappa,1)$, for example, then cardinals are preserved. I am interested to know if this holds in general.

Does $Add(\kappa,1)^L$ ever collapse cardinals?

• What is the partial order for $Add(\kappa, 1)$? – Nate Ackerman May 22 '15 at 4:25
• @Nate: It's the forcing $\operatorname{Add}(\kappa,1)$ (all the partial functions from $\kappa\to2$ with domain bounded below $\kappa$; ordered by reverse inclusion), but only those functions which are in $L$. – Asaf Karagila May 22 '15 at 6:01
• (I removed the [cohen-forcing] tag, I think it's overly specific.) – Asaf Karagila May 22 '15 at 6:02
• @Asaf: Thanks. But then this raises three other questions. First, why is this $Add(\kappa, 1)$ and not $Add(\kappa, 2)$? What is $Add(\gamma, \kappa^+)$? And then what is the difference between $Add(\kappa, 1)$ and $Add(\kappa, 1)^L$ (which is mentioned in the question as separate)? – Nate Ackerman May 22 '15 at 7:55
• @NateAckerman Adding one Cohen subset to $\kappa$ is isomorphic to adding two such sets, so $\text{Add}(\kappa,1)\cong\text{Add}(\kappa,2)$. In general, $\text{Add}(\kappa,I)$ should consist of partial functions on $\kappa\times I\to 2$, with support of size $\kappa$. In the case of $1$, we simply suppress the extra coordinate as unnecessary. The superscript $L$ on the notation $\text{Add}(\kappa,1)^L$ means that Jonas is considering the version of the forcing notion as defined in $L$ (by taking only such conditions that are in $L$), but he wants to force with that poset overe $V$. – Joel David Hamkins May 22 '15 at 10:52

Yes, it follows from a theorem of Mack Stanley (for this special case, in fact from a theorem of Foreman-Magidor-Shelah) that if $0^\sharp$ exists, then forcing with $Add(\kappa, 1)^L$ collapses $\kappa$ into $\omega.$
See Stanley's paper "Forcing disabled" (and Foreman-Magidor-Shelah paper "$0^\sharp$ and some forcing principles", J Symbolic Logic 51 (1986), 39-46). See also Can the Cohen forcing collapse cardinals? and the answer given there.