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I am trying to wrap my head around the application of the downward Löwenheim-Skolen theorem to ZF set theory. So, there is a countable model $(V,\in)$ of ZF set theory, where $V$ is a proper class of sets (the universe of sets). Ok, so what exactly is countable here? The class $V$ is proper, so it is not a set itself, therefore as I understand $V$ cannot be countable since a bijection $f: \mathbb{N}\rightarrow V$ doesn't exist. So my first question is:

$\quad (1)$ What exactly is countable in a countable model of ZF set theory?

Secondly, things like the power-set of $\omega$ ($\cong \mathbb{R}$) is uncountable, and yet it exists in a countable model? Is there a contradiction or issue with definitions here? Or is it just that uncountable sets can exist within a countable class of sets? So, second bulleted question:

$\quad (2)$ How can $\mathcal{P}(\omega)$ (or other uncountable sets) exist within a countable model of ZF set theory?

I apologize for the chaotic and verbal description of my thoughts, but this is more like a plead for help to improve my understanding rather than a formal proof-check or something of that nature. Thanks in advance.

Auclair
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Your confusion is that there are two different notions of "set" under discussion. Each notion of "set" comes with its own notion of "countable". These need not agree!!!

The "external" notion of set is the one given by the ambient language you're using to formulate logic and model theory. The "internal" notion of set is the one interpreted in the model of the formal theory under discussion.

Externally — i.e. using the relevant set theory coming from the ambient language — $V$ is merely countable. But internally — i.e. using the interpretation of the formal language you're modeling — $V$ is indeed a proper class.


For more reading, look for material on Skolem's paradox. However, take care; my past impressions from google search is that you see far too much weight placed on older reactions coming from the time when formal logic was a brand new subject that mathematicians were struggling to understand.

  • Hm, ok. So $\textit{externally}$, $V$ is a set which is countable. But within ZF set theory, it is a proper class? In any case, what is the notion of countable in the $\textit{internal}$ notion? – Auclair Jun 01 '18 at 04:03
  • @Auclair: The usual definition: that there exists a bijection with the natural numbers. You'll find that the model does not contain any set that is a bijection between the model's natural numbers and the class of all sets of the model. –  Jun 01 '18 at 04:09
  • Great. I think I'm beginning to get this now. I can see why it's called a paradox. Thank you for the help. – Auclair Jun 01 '18 at 04:11
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So, there is a countable model $(V,\in)$ of ZF set theory, where $V$ is a proper class of sets (the universe of sets).

That's not correct. A countable model $(V,\in_V)$ of ZF set theory is a model where $V$ is a countable set. So, therefore:

What exactly is countable in a countable model of ZF set theory?

The thing that's countable is the set $V$.

How can $\mathcal{P}(\omega)$ (or other uncountable sets) exist within a countable model of ZF set theory?

Well, there are elements $s$ of $V$ that satisfy the predicate "$s$ is uncountable", inside of the model. We can say that these elements $s$ are "internally uncountable".

This doesn't mean that the extension of $s$, $\{t : t \in V, t \in_V s\}$, is an uncountable set. It can't be an uncountable set, since it's a subset of $V$, which is a countable set.

What it does mean is that there's no element $f$ of $V$ which represents an enumeration of $s$. There is, in fact, an enumeration of $\{t : t \in V, t \in_V s\}$, but there is no element of the model which corresponds to this enumeration.

Tanner Swett
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  • Thank you for the clarification. So, let me check if I understand your answer. Say that $s$ is a countably infinite set inside the countable model $V$, then an enumeration (listing?) of $s$ would need to exist as an element of that model? And therefore, if an uncountable set exists inside that model, the enumeration by definition does not exist in the model? – Auclair Jun 01 '18 at 03:34
  • @Auclair Yes, that's all correct. – Tanner Swett Jun 01 '18 at 12:41