Timeline for Forcing conflation for $L(\mathbb{R})$
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Feb 1, 2017 at 2:22 | vote | accept | Noah Schweber | ||
| Feb 1, 2017 at 2:22 | comment | added | Noah Schweber | Sorry for the late reply - this is an excellent answer, and I yet again took something false for granted! Thank you very much! | |
| Jan 17, 2017 at 0:52 | comment | added | Todd Trimble | @JoelDavidHamkins It's not so simple as that; merging is done at the SE Community Management level, and more and more they ask the parties concerned (and not us moderators) to reach them directly. To user102684: if you go to MO meta and scroll to the bottom of the page, you will find 'Contact Us'. Hit that, and you will find a drop-down menu which includes merging. Describe your problem in the webform provided. Let me know here if you experience difficulty with this. | |
| Jan 17, 2017 at 0:26 | comment | added | Gabe Goldberg | When I said in my first comment that "the two models are equal" I meant the two models $L(\mathbb R)^{K[c]}$ and $L(\mathbb R)^K[c]$. (I merged the accounts, by the way, but they were both of the form "user $n$".) | |
| Jan 16, 2017 at 22:40 | comment | added | Joel David Hamkins | Thanks very much. The moderators can merge your accounts, if you just let them know the usernames. | |
| Jan 16, 2017 at 22:33 | comment | added | Gabe Goldberg | The statement "$L(\mathbb R)^{V[c]}\subsetneq L(\mathbb R)[c]$" has no consistency strength beyond ZFC plus an inaccessible. Suppose $G$ is $V$-generic for $\text{Col}(\omega,{<}\kappa)$. We have $\text{Col}(\omega,{<}\kappa) \simeq \text{Col}(\omega,{<}\kappa)\times \text{Add}(\omega,1)$. So by Solovay's theorem, $V[G]$ and $V[G][c]$ both satisfy the statement "Every uncountable set of reals in $L(\mathbb R)$ has a perfect subset," and this suffices to run my argument. Thanks for welcoming me, although actually I just accidentally created a second account. | |
| Jan 16, 2017 at 22:30 | comment | added | Gabe Goldberg | Suppose there is no inner model with a Woodin cardinal. Let $K$ denote the Jensen-Steel core model, and suppose $c$ is Cohen over $K$. One theorem of Steel implies that $K^{K[c]} = K$. Another states that $K\cap \textit{HC}$ is definable in $L_{\omega_1}(\mathbb R)$. Thus $K\cap \textit{HC} = K^{K[c]}\cap \textit{HC}^{K[c]}\in L(\mathbb R)^{K[c]}$. It follows that $L(\mathbb R)^K\subseteq L(\mathbb R)^{K[c]}$, and so the two models are equal. So in some sense no large cardinal hypothesis below a Woodin suffices to show $L(\mathbb R)^{L(\mathbb R)[c]}\subsetneq L(\mathbb R)[c]$. | |
| Jan 16, 2017 at 12:04 | comment | added | Joel David Hamkins | Welcome to MathOverflow! And thank you for this clarifying post. I wonder whether one can prove $L(\mathbb{R})^{V[c]}\subsetneq L(\mathbb{R})[c]$ under considerably weaker assumptions? | |
| Jan 16, 2017 at 11:18 | review | First posts | |||
| Jan 16, 2017 at 11:36 | |||||
| Jan 16, 2017 at 11:15 | history | answered | Gabe Goldberg | CC BY-SA 3.0 |