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I know that in ZFC that some collections of objects cannot be gathered together into a set (for example, the "set of all sets") does not exist, nor is "the set that just contains itself."

Is it possible to construct the set of all sets with cardinality $\aleph_0$? Or does this set not exist?

Thanks!

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4 Answers 4

up vote 8 down vote accepted

It does not exist. Suppose that it did, and call it $C$. The union axiom then ensures that the set $\bigcup C$ exists. But every set $x$ is an element of a countably infinite set, e.g., $\{x\}\cup\omega$, so for each set $x$ there is a $y\in C$ such that $x\in y$, and therefore $x\in\bigcup C$. That is, every set is an element of $\bigcup C$, which you already know is impossible. Thus, $C$ cannot exist.

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Beautiful, thanks! –  templatetypedef Jan 5 '13 at 22:57
    
@templatetypedef: My pleasure! –  Brian M. Scott Jan 5 '13 at 22:58

No.

The reason is simple:

For every set $A$, the set $\{ A \} \cup \mathbb N$ is countable. And the set of all such sets would contain the "set" of all sets as a subset.

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The answer is ‘no’, as shown by many people here.

There are two ways of trying to make sense of the ‘set of all sets’, although not in the literal sense. If you are looking for a model of ZFC, then you can consider the collection of all sets whose rank is less than some inaccessible cardinal (assuming that one exists at all). This collection is, of course, still a set in the universe, and it yields a standard model of ZFC. However, inaccessible cardinals are uncountable, so this approach does not yield a countable ‘set of all sets’. Note the quotation marks!

However, if ZFC is consistent, then there exists a countable model of ZFC, which is a consequence of the Löwenheim-Skolem Theorem from logic and model theory. Therefore, you may want to take this model to be your ‘set of all sets’.

Whatever it is, from the model-theoretic point of view, this ‘set of all sets’ is never going to be an element of itself.

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Suppose $x$ is a set, and consider corresponding set

$$X_x = \{(0, X), (1, X), (2, X), \ldots \}$$

Then each $X_x$ has cardinality $\aleph_0$, and there are as many of the $X_x$ as there are of sets $x$, i.e. too many to form a set.

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