1. Is ZFC2U categorical? This is second-order ZFC with full semantics and urelements.

  2. Is the cardinality of a set A from ZFC2U expressible by a sentence in second order logic?

In 1, because there is not a syntactical deductive recursive calculus for 2nd order logic, I don´t know how to try the demonstration.

In 2, I know that the uncountability of a set says that there exist two infinite subsets of the set that are not bijectable. I suppose that it can be to say a generalization of it, but I don´t know how to continue.

  • $\begingroup$ @Jesse What is the reason why doy you think my question is unclear. Maybe is my english not good? It could be, I think. I greatly appreciate the answer because I want still be able to ask questions and collaborating forum $\endgroup$ – Raúl Aparicio Bustillo Nov 17 '15 at 15:47
  • $\begingroup$ questions without the OP's (the questioner) thoughts/attempt to solve it etc. are always discouraged here. Adding your attempt "helps others identify where you have difficulties and helps them write answers appropriate to your experience level". $\endgroup$ – Jesse P Francis Nov 17 '15 at 17:06
  • $\begingroup$ Message for OPs: I wish delete this question, because I would fuse it with my previous question about categoricity of ZFC2. Could it be possible? Or it is better letting it be because there are some information about how to answer these questions. I don´t wanna that it look like a repeated question. And I would like the concrete reasons my account was blocked to ask more times. My graduate formation is scientifical but no mathematical direct, but I am very interested knowing some questions about foundations of mathematics. Sorry for my bad english, I am not speaker of it. $\endgroup$ – Raúl Aparicio Bustillo Nov 18 '15 at 7:16
  • $\begingroup$ @Jessie ¿How can Iproceed to solicitate my admission for asking new questions? $\endgroup$ – Raúl Aparicio Bustillo Nov 18 '15 at 10:39
  • $\begingroup$ @Jessie I have no reputation to talk in this link, Another way for communicate with OPs? $\endgroup$ – Raúl Aparicio Bustillo Nov 18 '15 at 13:07

No - under reasonable assumptions, second-order ZFC is not categorical (passing to ZFCU makes things even less categorical), and there are cardinalities which are not second-order definable, in the sense that there are sets $X, Y$ of different cardinality such that $X$ and $Y$ satisfy the same second-order sentences.

Let's tackle the second point first. There are only countably many second-order sentences, so there are only continuum-many complete second-order theories in the language $\{A, \in\}$ - so, since there are more than continuum-many cardinalities, there must be two sets $X$ and $Y$ of different cardinality such that $(V; X, \in)$ and $(V; Y, \in)$ are second-order elementarily equivalent. (Here "$V$" denotes the universe of sets.) This has plenty of obvious variations.

Now, my pigeonhole argument relies on some assumptions about the universe of sets which are both "obvious" (I tend to think) and also not consequences of ZFC alone - for instance, that $V$ is not pointwise definable. But I think it would be more pedagogically useful for you to first understand why the argument above is plausible, and then to pick apart at its additional assumptions.

(By the way, you may be interested in the concept of Hanf numbers.)

As to the categoricality of second-order ZFC: well, this is more complicated, but has been asked before. See Why isn't second-order ZFC categorical?.

  • $\begingroup$ ¿Are you edited the response? I was wathing the answer an hour ago and I looked that ZFC+ there are no inaccesible cardinals was categorical $\endgroup$ – Raúl Aparicio Bustillo Nov 16 '15 at 8:54
  • $\begingroup$ You are correct, "(second-order) ZFC + no inaccessibles" is categorical - but that's a different question. And most of the interest in extensions of ZFC right now focuses on having more large cardinals, not fewer (there are of course exceptions). $\endgroup$ – Noah Schweber Nov 16 '15 at 8:58
  • $\begingroup$ And what´s the matter with NBG+urelements+no accesibles. Is it categorical too? $\endgroup$ – Raúl Aparicio Bustillo Nov 17 '15 at 15:43
  • 2
    $\begingroup$ I don't understand what urelements are doing here - as soon as we allow them, we're definitely not categorical, since e.g. the statement "There are exactly seven urelements" is left undecided by the theory. As to second-order NBG+no inaccessibles, I'm not sure, but I suspect this is categorical - certainly second-order ZFC + no inaccessibles is categorical, since the models of second-order ZFC are exactly the sets $V_\kappa$ for $\kappa$ an inaccessible. $\endgroup$ – Noah Schweber Nov 18 '15 at 7:11

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