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Define $C=\{\alpha:\alpha=|x|$ for some set $x$$\}$ as the class of all cardinals. ($|x|$ being the cardinality of the set $x$)

It will be enough to prove $C$ is a proper class by showing $On\subseteq C$

Since we can take it to be an established fact that $On$ is a proper class.

Let $\alpha \in On$. Assume $\alpha \notin C$

So $\neg\exists x \in V $(the universal class) such that $|x|=\alpha \implies$

So $\neg\exists x \in V $ st. $\alpha$ is the least ordinal such that $\alpha \approx x$

But it's trivially true that $\alpha \approx \alpha \in On \subseteq V$. So $\alpha \in V$. This would be a contradiction since we assumed there is no such set.

Therefore, $\alpha \in C \implies On\subseteq C$ and then $C$ is a proper class.

$\textbf{Edit:}$ As pointed out by Pedro, the initial assumption of $On\subseteq C$ is false. In fact, $C \subseteq On$ is true - since all cardinals are necessarily ordinals.

Andrés E. Caicedo
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Walt van Amstel
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    It is false that $On\subseteq C$. For instance, the ordinal $\omega+1 = {0,1,2\dots,\omega}$ is not a cardinal: The least $\alpha$ bijective with $\omega+1$ is $\omega$. The bijection is given by the following enumeration: $\omega,0,1,2,\dots$. – Pedro Sánchez Terraf Mar 25 '16 at 16:15
  • @Pedro Sánchez Terraf: Thank you for the response. Would you be so kind to suggest a new starting point ? An obvious one would be assuming for contradiction that $C$ is indeed a set. – Walt van Amstel Mar 25 '16 at 16:27
  • Consider $\aleph_\alpha$. – Wojowu Mar 25 '16 at 16:47
  • For the coursework I am doing, this question precedes any mention of alephs. – Walt van Amstel Mar 25 '16 at 16:52
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    You need at least to know that for every set $A$ there is a cardinal $\kappa$ that does not inject into $A$. Given this, if $C$ were a set, then $\bigcup C$ would be a cardinal, and then get a contradiction. – Pedro Sánchez Terraf Mar 25 '16 at 17:25
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    Regarding the hint in the last comment, this is easy if you are assuming the axiom of choice. If not, you need an additional argument to show that for any (well-orderable) cardinal $\kappa$ there is a larger such cardinal. – Andrés E. Caicedo Mar 25 '16 at 17:37

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It turns out it would be enough to prove $On \subseteq \bigcup C$:

Suppose $C$ is a set. Then we know $\bigcup C$ is also a set from the unions axiom. If the above claim holds true then it would follow that $On$ is a set $-$ which is known not to be true.

Let $\alpha \in On$

A previously proven theorem I will assume is Cantor's Theorem:

For any set $x$, we know $x\prec p(x)$

We know $\alpha$ is a set since all ordinals are sets.

It follows $\alpha \prec p(\alpha)$

Every set is bijectively equivalent to its own cardinal so $p(\alpha)\approx |p(\alpha)|\in On$

The least ordinal that satisfies the property $p(\alpha)\approx \beta$ is the cardinal of $p(\alpha)$, we know at least one such ordinal exists by the Numeration Principle.

So, $\alpha \prec p(\alpha)\approx|p(\alpha)|\in On $

It has also been shown previously if $\alpha \prec p(\alpha)$ $\implies |\alpha| \lt |p(\alpha)|$, but $|\alpha|\approx \alpha$

So $\alpha \lt |p(\alpha)|\implies \alpha \in |p(\alpha)|$ since $\alpha, |p(\alpha)|\in On $

Now, $ p(\alpha)$ is a set by the power set axiom. So we can deduce from the definition of $C$ that $| p(\alpha)|\in C\implies \alpha \in \bigcup C$

So, it holds true that $On\subseteq \bigcup C$, and we have the necessary contradiction.

Walt van Amstel
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  • What do you mean by $\bigcup C$, $\prec$, and $p(x)$? Is $p(x)$ a property of $x$? You never introduced a property $p$ before you wrote $p(x)$. Is $p(x)$ the power set of $x$? – Timothy Jan 11 '18 at 20:02
  • $\bigcup C$ means the elements of all elements of $C$. $A \prec B$ means that $A$ is strictly embeddable in $B$ and $p(x)$ is indeed the powerset of $x$. – Walt van Amstel Jan 23 '18 at 06:28
  • Are you using $C$ to mean the collection of all cardinal numbers? Is $\biggup C$ the union of all elements of C? I only know of an ordinal number being a set and not a cardinal. – Timothy Jan 23 '18 at 19:35