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

Question

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 $|\omega|=|\omega+1|$?

Answer 1

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

Answer 2

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 $$0 \preceq 1 \preceq 3 \preceq 5 \preceq \cdots \preceq 2 \preceq 6 \preceq 10 \preceq \cdots \preceq 4 \preceq 12 \preceq 20 \preceq \cdots $$

Answer 3

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.