# Why $\omega+1$ and $\omega^2$ are not cardinal numbers?

I see the following definition of cardinal number in notes:

An ordinal $\alpha$ is a cardinal number if $|\beta|<|\alpha|$ for all $\beta\in\alpha$.

Why $\omega+1$ and $\omega^2$ are not cardinal numbers?

For $\omega+1$, is it because $\omega\in\omega+1$ but $|\omega|=|\omega+1|$?

## 3 Answers

It is precisely because $| \omega | = | \omega + 1 | = | \omega^2 |$ (and $\omega < \omega+1 < \omega^2$). Perhaps to make this clear, we can define well-orderings on $\mathbb{N}$ with these prescribed order-types (meaning that the underlying set of any well-ordering with these order-types is countably infinite; i.e., equinumerous to $\omega$):

• $\omega + 1$ is the order type of the following well-ordering on $\mathbb{N}$: $$m \preceq n \Leftrightarrow \begin{cases} 0 < m \leq n, &\text{or} \\ n = 0 \end{cases}$$ So this well-order looks like $$1 \preceq 2 \preceq 3 \preceq \cdots \preceq 0$$

• $\omega^2$ is the order-type of the following well-ordering on $\mathbb{N}$: $$m \preceq n \Leftrightarrow \begin{cases} m = 0, &\text{or} \\ ( \forall i ) ( 2^i \mid m \leftrightarrow 2^i \mid n ) \wedge m \leq n, &\text{or} \\ ( \exists i ) ( 2^i \not\mid m \wedge 2^i \mid n) \wedge n \neq 0 \end{cases}$$ So this well-order looks like $$0 \preceq 1 \preceq 3 \preceq 5 \preceq \cdots \preceq 2 \preceq 6 \preceq 10 \preceq \cdots \preceq 4 \preceq 12 \preceq 20 \preceq \cdots$$

Indeed, we have $\omega\in\omega+1$ and $\omega\in\omega^2,$ but $|\omega+1|=|\omega|$ and $|\omega^2|=|\omega|^2=|\omega|.$

• Why $|\omega^2|=|\omega|^2$? Could you prove it? – setvomega Dec 4 '13 at 18:43
• Certainly. What is your definition of ordinal exponentiation? – Cameron Buie Dec 4 '13 at 18:47
• $\omega^2=\omega\cdot\omega$ is the unique ordinal isomorphic to the lexicographic order of $\omega\times\omega$. – setvomega Dec 4 '13 at 19:29
• By the same token, $|\omega|^2$ is the unique cardinal in bijection with $|\omega|\times|\omega|.$ Now, if $A$ and $B$ are sets of (respective) cardinality $\kappa,\lambda,$ then letting $f:A\to\kappa$ and $g:B\to\lambda$ be bijections, we can readily define a bijection $h:A\times B\to\kappa\times\lambda$ by $$h(a,b)=\bigl(f(a),g(b)\bigr),$$ so $$|A\times B|=\kappa\times\lambda=|A|\times|B|$$ in general. In particular, when $A=B=\omega,$ we find that $|\omega^2|=|\omega|^2.$ – Cameron Buie Dec 4 '13 at 21:35

Remember these two theorems:

1. The union of two countable sets is countable.
2. The product of two countable sets is countable.

Now we have that $\omega+1$ is the union of a countable set with a finite set; and $\omega^2$ is order isomorphic to the product $\omega\times\omega$ with the lexicographic order. Therefore the underlying sets of both these ordinals are countable.