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Is there a universe which can always be forced to, which never can be forced from?

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No; given any universe, we can force either $CH$ or $\neg CH$, and clearly one of these must result in a universe with different properties. – Noah Schweber Mar 20 '14 at 4:32
It would also have to be universe to which every universe can force to. – Erin Carmody Mar 20 '14 at 4:33
Looks cool, I'll check out the link. Did you have any mathematical comments to make on this site? – Erin Carmody Mar 20 '14 at 4:45
In the context of the multiverse, you can force from any model of $\mathsf{KP}$, in particular from any model of set theory. If $M$ is a model, no proper forcing extension of $M$ equals $M$, so the answer to the question is no. – Andrés E. Caicedo Mar 20 '14 at 6:05
As an aside, if you are instead interested in specific statements which have this property, then these are what are known as buttons in the area of the modal logic of forcing. Examples are easy to find depending on which direction of the multiverse you are interested in ($\mathsf{V=L}$, $\mathsf{V\neq L}$, contains a real Cohen over $L$ etc), but there are some interesting questions (raised by Joel Hamkins and Benedikt Loewe) concerning the number and the interrelations of the buttons of models of $\mathsf{ZFC}$. – tci Mar 20 '14 at 11:35
up vote 7 down vote accepted

I think there's a serious confusion going on here, around what sort of background we assume.

If we are working within a single model $V$ of $ZFC$, and considering the multiverse(-like structure) consisting of all the inner models of $V$, then there are lots of possibilities. For one thing, if we include $V$ here, there is obviously a 'terminus.' However, even if we only include (say) generic extensions within $V$ of $L$, there are still lots of possibilities; for example, Monroe's example of a model with precisely two inner models: $L$ and itself. So in this case, there are lots of possibile answers, depending on exactly how you set things up.

If, however, we adopt a truly multiversal picture of things, then at the very least we want to demand that we can always force - in every presentation of the multivierse or anything similar I've ever seen, this has been one of the axioms. In this case, it's trivially the case that there is no terminus, since by definition we can always force over any world in the multiverse.

Where you're getting confused, I suspect, is in mixing the two pictures - talking about multiversal ideas (such as set-theoretic geology) while holding a picture of an 'ultimate' $V$ in the background, which of course constrains the allowed forcings. I suggest you make very precise exactly what your background theory is here, what sort of things we're assuming, etc., because otherwise this question is just too broad to admit a real answer.

One particular nice and specific sub-question lurking here: what are the possibilities for "all the set-generic extensions of $L$?" (Within a fixed $V$.) This could be a singleton, or have a top element, or be just a hideously complicated directed system of things! If what you're really interested in is the sort of variety 'the inner models of $V$' can have, I suggest among other things looking at Sy Friedman's Inner Model Hypothesis (

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Really really interesting Noah. You might have solved the question. My inner model is ultimate L while I roam the multiverse. So, I guess there can't be one universe in the multiverse which fulfills the role of ultimate L, unless there is another answer. – Erin Carmody Mar 20 '14 at 6:34
I don't quite understand what you mean by "my inner model." Are you picturing your collection of viable universes as being outer models of ultimate $L$, or what? (Also, what exactly do you mean by "the role of ultimate $L$?" I sort of understand what it's doing within a single $V$, but I don't quite understand what you're getting at.) – Noah Schweber Mar 20 '14 at 6:46
I am gaining an intuition by imagining the universes as real, so it might be better to take it softly, as it is intended. – Erin Carmody Mar 20 '14 at 7:18

Is there a terminus? That'd be a universe you can't "force from." If you can always take a generic extension, then there's no terminus.

Let's be real. You can't really force away from the universe. It's the universe; it's all there is. It's the last train station. Is it consistent that every other train stop goes to Terminus?

Yes, look at Sacks forcing. Add a Sacks real $s$ to $L$. It's a fact (see Jech) that there are no proper intermediate models between the ground model and the Sacks extension. So if we live in $L[s]$, there are only two inner models: $L$ and $L[s]$. So there's only one route, all aboard!

Now what counts as a universe? Is it only a ZFC model? If so, any forcing over $L$ gets a similar situation. In general, every intermediate submodel of ZFC of a forcing extension containing the ground model is a forcing extension. (This is a really great fact, see Jech again.) So force over $L$ with whatever, and again we get that every train station has a route to the Terminus.

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That's a good one Monroe. Every universe which can be forced to is different from every other universe. So the universe cannot be described by an extension. – Erin Carmody Mar 20 '14 at 5:11
This is a nice answer! But, since the setting of the question is set-theoretic geology, I presume the "background" reality is that we can always take forcing extensions. – Noah Schweber Mar 20 '14 at 5:11
Erin, I don't understand either your comment, or how it follows from Monroe's answer; could you explain? – Noah Schweber Mar 20 '14 at 5:12
You say that there is no universe from which all extensions can be reached by forcing. And an earlier comment pointed out that we can always force. So, I think that's a full answer if you can show that every universe can force. – Erin Carmody Mar 20 '14 at 5:16
Consistently, every inner model of ZFC is a forcing extension of L... – Monroe Eskew Mar 20 '14 at 5:20

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