Continuum hypothesis quickfur Junior Member Posts: 32 Threads: 9 Joined: Dec 2006 Reputation: 0 2006-12-13, 00:34 This is really more a question than a real idea... we know that the continuum hypothesis $2^{\aleph_0}=\aleph_1$ is undecidable in ZFC. How general is this result? Does the undecidability also include $2^{\aleph_0}=\aleph_i$ for all finite i? All transfinite i? I wonder if there has been any research in the direction of $2^{\aleph_0} = \epsilon$ where $\epsilon = \aleph_{\epsilon}$. Just curious. I'm no expert in large cardinals/ordinals, but it is a rather fascinating subject. thigle Newbie Posts: 1 Threads: 0 Joined: Dec 2006 Reputation: 0 2006-12-15, 18:10 hi quickfur. how do we know that the continuum hypothesis 2N0=N1 (sorry i don't know how to enter those symbols) is just that ? what does it mean? please can you explain this hypothesis a bit ? or where should i look into this matter ? and what is ZFC ? bo198214 Administrator Posts: 53 Threads: 7 Joined: Dec 2006 Reputation: 0 2006-12-15, 18:34 Hello thigle! Welcome on board! (btw. you have switched off your PM, was this intentionally?) thigle Wrote:2N0=N1 (sorry i don't know how to enter those symbols)If you simply want to write a symbol/formula, you have seen somewhere, you can hit the reply button, and then see the tex-source. In this case Code:$$2^{\aleph_0}=\aleph_1$$ Quote:what does it mean? please can you explain this hypothesis a bit ? or where should i look into this matter ?and what is ZFC ? Oh thigle, then you have quite a bit studying before you. But it is indeed an interesting topic, and touches the foundations of mathematics. ZFC is very good explained at wikipedia, as well as is the continuum hypothesis. So, be indulgent with me not explaining it myself. After your reading we awaiting your questions I indeed will add a questions section and move this thread thereto. quickfur Junior Member Posts: 32 Threads: 9 Joined: Dec 2006 Reputation: 0 2006-12-15, 20:34 bo198214 Wrote:[...]I indeed will add a questions section and move this thread thereto.Thanks. Do you have any answers to this topic, though? bo198214 Administrator Posts: 53 Threads: 7 Joined: Dec 2006 Reputation: 0 2006-12-15, 20:56 quickfur Wrote:Do you have any answers to this topic, though? DirkU should be the set theory specialist, but he is currently on vacation, and my knowledge of set theory is probably even less than yours ... DirkU Junior Member Posts: 8 Threads: 1 Joined: Dec 2006 Reputation: 0 2007-01-10, 11:20 bo198214 Wrote:... DirkU should be the set theory specialist, but he is currently on vacation, and my knowledge of set theory is probably even less than yours ... How nice to be called a set theory expert ... To say something to the question (as I understood it) I have first something to say to a method of set-theoretical independence proofs. That $2^{\aleph_0}=\aleph_1$ is not provable in ZF is usually shown by a forcing construction. There are some variants to do so. Often you will start with a ZF model $\mathcal{M}$ and construct another one--say $\mathcal{M}^*$ -- from $\mathcal{M}$ which is forced to have some desired properties--for instance that the negation $\neg\phi$ holds in $\mathcal{M}^*$ of a statement $\phi$ you want to prove to be independent of ZF. If this is possible then $\neg\phi$ cannot proved in ZF. Now the standard construction of the model $\mathcal{M}^*$ to disprove $2^{\aleph_0}=\aleph_1$ in ZF is done in way to move $2^{\aleph_0}$ upwards in the hierarchy of cardinal numbers. When you analyze this construction you can see that you can make $2^{\aleph_0}$ make nearly arbitrary large, i.e. you can choose an ordinal number $\alpha$ much greater than 2 and force that $2^{\aleph_0} = \aleph_{\alpha}$ in $\mathcal{M}^*$. Thus for every ordinal number $\beta$ such that another such forceable ordinal number $\alpha$ as just described with $\beta > \alpha$ exists it is unprovable in ZF that $2^{\aleph_0} = \aleph_{\beta}$. « Next Oldest | Next Newest »