The consistency of ZFC + CH gives the ability to travel to a universe which models ZFC + \neg CH? - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-06-19T22:26:07Z http://mathoverflow.net/feeds/question/43362 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43362/the-consistency-of-zfc-ch-gives-the-ability-to-travel-to-a-universe-which-model The consistency of ZFC + CH gives the ability to travel to a universe which models ZFC + \neg CH? Erin K Carmody 2010-10-24T07:46:12Z 2012-03-13T01:54:15Z <p>Why does forcing seem to be so vacuously true?</p> <p>It seems like you are just reversing the subset containment of the model of ZFC + CH to be the other way in the poset. So, why is this valid? Why are you allowed to just put the empty set at the top of the partial order*?</p> <p>*I have just started learning set theory so I have only seen one example of forcing and it was the forcing of the set of reals to be equal to the second uncountable cardinal.</p> http://mathoverflow.net/questions/43362/the-consistency-of-zfc-ch-gives-the-ability-to-travel-to-a-universe-which-model/43365#43365 Answer by Bjørn Kjos-Hanssen for The consistency of ZFC + CH gives the ability to travel to a universe which models ZFC + \neg CH? Bjørn Kjos-Hanssen 2010-10-24T08:28:41Z 2010-10-24T08:28:41Z <p>The empty set $\emptyset$ here is the forcing condition that has the least amount of information about the generic object being constructed. </p> <p>Since it has the least information, you might think it should be at the bottom of the partial order. There are two replies to this: </p> <p>(1) Mathematically, we are free to define our partial order however we want, even if it confuses the reader.</p> <p>(2) In this case, there is no intent to confuse the reader. Having the <em>least</em> information, the condition $\emptyset$ also leaves the <em>greatest</em> amount of possibilities still open. So it is really a 50-50 decision which way one should define it. (Others may know the history of this particular convention.)</p> <p>A similar convention question is whether the Turing degree of the computable sets, $\mathbf 0$, is the smallest or the largest Turing degree. Conventionally one speaks of the degrees of <em>unsolvability</em> and so $\mathbf 0$ is the smallest, but if the convention were degrees of <em>solvability</em> it would be the greatest. Since none of the other degrees are very solvable at all, the current convention is perhaps the best in this case.</p> http://mathoverflow.net/questions/43362/the-consistency-of-zfc-ch-gives-the-ability-to-travel-to-a-universe-which-model/43370#43370 Answer by Stefan Geschke for The consistency of ZFC + CH gives the ability to travel to a universe which models ZFC + \neg CH? Stefan Geschke 2010-10-24T10:24:16Z 2010-10-24T10:41:14Z <p>Bjorn already mentioned one important point: We can define a partial order in any way we want. It should also be mentioned that Cohen defined the order in the other direction: He considered the empty condition the smallest and defined everything else accordingly (i.e., just the other way around). Shelah for instance still uses Cohen's original version of forcing. </p> <p>Considering $\emptyset$ the largest condition in this particular forcing context agrees with the usual order on Boolean algebras in the following way:<br> When forcing with Boolean algebras, the elements of the Boolean algebra correspond to truth values of sentences. If a sentence implies another, the truth value of the first sentence is less or equal to the truth value of the second sentence. The empty condition in Cohen forcing is the truth value of all sentences that are true in every forcing extension by Cohen forcing. This is the largest possible truth value in this context. (This is not to be confused with the fact that $\emptyset$ is the smallest element of the Boolean algebra $\mathcal P(\omega)$. The Boolean algebra corresponding to Cohen forcing is very different from $\mathcal P(\omega)$.) This interpretation of the direction of the order agrees with Bjorn's explanation saying that $\emptyset$ leaves open the most possibilities.</p> <p>Anyhow. The whole discussion above in somewhat unnecessary since your problem is actually at a different level: The partial order that we are talking about is just a technical tool in order to construct a forcing extension. It has little to do with the containment relation on the subsets of $\mathbb N$. In fact, the containment relation among subsets is the same in the ground model (the one that satisfies CH) and the generic extension. I.e., if $a,b\subseteq\mathbb N$ are both in the ground model, then $a\subseteq b$ holds in the extension iff it holds in the ground model. In the extension we just have many more subsets of $\mathbb N$. </p>