Timeline for special extremally disconnected spaces with only finite isolated points
Current License: CC BY-SA 3.0
20 events
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| Apr 13, 2019 at 11:45 | review | Close votes | |||
| Apr 13, 2019 at 20:45 | |||||
| Nov 22, 2012 at 23:45 | history | edited | Goldstern |
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| Nov 22, 2012 at 23:31 | answer | added | Goldstern | timeline score: 8 | |
| Nov 22, 2012 at 23:25 | comment | added | Goldstern | I think the "well-known theorem" is not formulated correctly. If you add to an extremally disconnected P-space any number of discrete points, the space is still extremally disconnected and a P-space. If the original space was not discrete, the new space won't be either. Perhaps you meant to say that such a space has to have AT LEAST the cardinality of the least measurable. | |
| Nov 22, 2012 at 23:01 | comment | added | Goldstern | You probably want your space to be Hausdorff, otherwise there is an example with 2 points. | |
| Nov 22, 2012 at 20:42 | comment | added | Goldstern | If there are only finitely many isolated points, you can remove them and are still left with an extremely disconnected space, right? And still a P-space? But now without isolated points. | |
| Nov 22, 2012 at 19:55 | history | edited | Ali Reza | CC BY-SA 3.0 |
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| Nov 22, 2012 at 19:51 | comment | added | Ali Reza | Dear Goldstern, I should recall that a P-space is compact if and only if it is finite. so it doesn't make sense in our problem. The reference for the well-Known theorem in my question is: exercise [12 H.6] of chapter 12 in the text "rings of continuous functions". | |
| Nov 22, 2012 at 19:25 | comment | added | Goldstern | Can you please give a reference for the "well-known theorem"? It might help those who want to learn more about the known background. Also, do you know if such spaces can be compact? If you are looking for compact (0-dim, which is true here) spaces, then we could translate the problem to a problem about Boolean algebra. | |
| Nov 22, 2012 at 19:24 | comment | added | Goldstern | Your title "On the existence of measurable cardinals" is extremally non-descriptive. I suggest "extremally disconnected spaces without isolated points" instead. | |
| Nov 22, 2012 at 17:45 | comment | added | Asaf Karagila♦ | Well, you probably know that you cannot prove the consistency of ZFC from the axioms of ZFC. The theory "ZFC+There exists a measurable cardinal" is a very strong theory compared to ZFC, in particular it proves the consistency of ZFC (if we assume a measurable exists then we can prove that ZFC is consistent). So assuming a measurable is a stronger assumption than assuming that there are no measurable cardinals. There is a lot of delicacy in those arguments, especially if you are unfamiliar with consistency results. But if there is a measurable, then there is a least measurable cardinal. | |
| Nov 22, 2012 at 17:24 | history | edited | Ali Reza | CC BY-SA 3.0 |
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| Nov 22, 2012 at 17:23 | comment | added | Ali Reza | @Asaf. you are right. I am not a set Theorist and since i am working in ZFC, i have thought that this problem is an open problem. I do not Know why I have thought that this is an unsettled problem. But there is another Question. you claim that this is unprovable. when we consider this, can we assume if there is a least measurable cardinal or there is not a measurable cardinal, if we work in "ZFC"? for notation i will fixe it as soon as possible. | |
| Nov 22, 2012 at 14:31 | comment | added | Asaf Karagila♦ | AliReza, the term "unsettled" usually means "an open problem", for example it is unsettled if the nontrivial zeros of the Riemann zeta function lie on the line $Re(z)=\frac12$. Unprovable is a form of settled. It is an answer. We cannot prove the existence of measurable cardinals, nor we can disprove it. | |
| Nov 22, 2012 at 13:43 | history | edited | Ali Reza | CC BY-SA 3.0 |
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| Nov 22, 2012 at 13:00 | comment | added | Qfwfq | (A minor point. I dont see the difference between finding "a cardinal number with measurable cardinal" and finding "a measurable cardinal". According to the definition you give, they're meant to be the same, right? ) | |
| Nov 22, 2012 at 12:16 | comment | added | Ali Reza | @ Asaf, I only assumed that "if" this cardinal exists. In this case it's not important for me to work in which Axiomatic system larger than "ZFC". for unsettled i mean unprovable in "ZFC" to | |
| Nov 22, 2012 at 12:15 | history | edited | Ali Reza | CC BY-SA 3.0 |
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| Nov 22, 2012 at 11:42 | comment | added | Asaf Karagila♦ | What do you mean it is unsettled? We know exactly that from the axioms of ZFC it is unprovable that a measurable exists. In fact from ZFC and many other large cardinal axioms it is unprovable that a measurable exists. However from assumptions like "There is a Woodin cardinal" or "There is a supercompact cardinal" or "AD holds in $L(\mathbb R)$" we can prove that a measurable exists. | |
| Nov 22, 2012 at 11:39 | history | asked | Ali Reza | CC BY-SA 3.0 |