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In ZF, not all "collections of objects" are sets. For example, there is no set of all sets, and there is no set of all ordinals.

So, how do we know that there is a set of all countable ordinals? In other words, how do we know that $\omega_1$ exists? (I'm assuming you don't need Choice.)

Akiva Weinberger
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  • There are other duplicates as well. Please search before posting in the future. – Asaf Karagila Feb 08 '15 at 21:54
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    @AsafKaragila Sorry. I searched for things related to $\omega_1$; I didn't think about searching for $\aleph_1$. – Akiva Weinberger Feb 08 '15 at 21:56
  • This also comes to mind (perhaps as a better duplicate, too (also looking at this nowadays makes me embarrassed for asking that question at the time! :-))) – Asaf Karagila Feb 08 '15 at 22:03
  • Of course it comes to your mind — you were the one who asked that! But I see your point; I didn't search as much as I should have. – Akiva Weinberger Feb 08 '15 at 22:05
  • @AsafKaragila Really quick question, slightly off-topic: What is $V=L$? Is it "A set exists iff we can prove that it exists?" – Akiva Weinberger Feb 08 '15 at 22:10
  • No. $V$ is the universe of sets, and you can construct it as a hierarchy where each step you take the power set of the previous one (and unions at limits), $L$ is Godel's constructible universe, each step you take the first-order definable subsets of the previous step (possibly with parameters). Both hierarchies give us models of $\sf ZF$, but $L$ will satisfy much more. It turns out there is a formula $\varphi(x)$ which is satisfied exactly by the sets which are in $L$. So $V=L$ means that all sets come from $L$. [...] – Asaf Karagila Feb 08 '15 at 22:36
  • [...] It turns out that $L$ is this "minimal" universe of set theory, in some very deep sense. So $V=L$ means that the universe is sort of minimal. But this minimality makes it very rich in provable combinatorial structure, and we can show that a lot of "good things" happen in $L$ (e.g. $\sf GCH$, all sort of guessing principles and so on and so on). – Asaf Karagila Feb 08 '15 at 22:37
  • Let us continue this discussion in chat. – Akiva Weinberger Feb 08 '15 at 23:22
  • Sorry, I don't like the chat here. But enjoy the rest of your evening. – Asaf Karagila Feb 08 '15 at 23:23
  • OK. You too, אסף. – Akiva Weinberger Feb 08 '15 at 23:24

1 Answers1

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Its a consequence of the axiom of replacement. The set of all binary relations on $\omega$ exists, its just $P(\omega \times \omega)$ and the subset $X$ of all well orders of $\omega$ also exists. To every well order of $\omega$ there is a well defined ordinal number, $\omega_1$ is the image of $X$ under the assignment of ordinal numbers.

Incidentally replacement is needed even to show that $\omega+\omega$ exists.

Rene Schipperus
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