If $1+ \alpha = \alpha$, $\alpha$ is an infinite ordinal

Question

I need to prove the following statement: If $1+ \alpha = \alpha$, $\alpha$ is an infinite ordinal.

I am trying to use Bernstein's Theorem(CBS) to show if $1+ \alpha \leq \alpha$, i.e., there is an injection from $1+ \alpha$ to $\alpha$, $\alpha$ must be an infinite ordinal.

Is this the right approach? I feel like this statement can be proven using a simpler approach like induction or etc., but I can't seem to think of one.

Also, does the converse always hold?

Thank you in advance.

Answer 1

Using Cantor–Bernstein is not the right tool here. You're not trying to prove that cardinalities are equal, but rather than the ordinals are equal. For this you need more than a bijection. You need an order isomorphism.

One way to simplify this is to remember that $1+\alpha=1+\omega+\beta$ for some $\beta$ such that $\omega+\beta=\alpha$, assuming that $\alpha$ is infinite.

So it is enough to show that for $\omega$, $1+\omega=\omega$. But that's fairly straightforward.

Answer 2

In the following, we will use the von Neumann definition of ordinal numbers.

First suppose that $1 + \alpha = \alpha$. This means that there is an order isomorphism $f: B \rightarrow \alpha$, where $B$ is the set

$B = \{(0, 0)\} \sqcup \alpha \times \{1\}$

imbued with an ordering $<$ defined by

$(0, 0) < (\beta, 1)$ for all $\beta \in \alpha$ and $(\beta, 1) < (\beta', 1)$ if and only if $\beta < \beta'$ for all $\beta, \beta' \in \alpha$.

To prove that $\alpha$ is infinite, we just need to construct an injection $h: \omega \rightarrow \alpha$. To this end, let $\iota: \alpha \rightarrow B$ denote the natural inclusion $\iota(\beta) = (\beta, 1)$, and let $g = f \circ \iota: \alpha \rightarrow \alpha$. Since $f$ and $\iota$ are both injective, so is $g$. Then we can define $h: \omega \rightarrow \alpha$ by $g^n(f(0, 0))$, where by convention $g^0 = \mathbb{1}_{\alpha}$ is the identity map.

To show that $h$ is injective, suppose for a contradiction that there are $m, n \in \omega$ such that $h(m) = h(n)$ but $m \neq n$. From the definition of $h$, this means $g^m(f(0, 0)) = g^n(f(0, 0))$. Assume without loss of generality that $m < n$. Then because $g$ is injective, $g^0(f(0, 0)) = g^{n - m}(f(0, 0))$. That is, $f(0, 0) = g^{n - m}(f(0, 0))$. But $f(0, 0)$ is not in the image of $g$, since $(0, 0)$ is not in the image of the inclusion $\iota$.

We conclude that $h$ is injective, hence $\alpha$ is infinite. Note that we did not have to use the fact that $f$ is an order isomorphism to prove that $\alpha$ is infinite, only that $f$ is injective.

The converse is also true. To see this, suppose that $\alpha$ is infinite. Then $\omega \subseteq \alpha$, so we need only find an order isomorphism $\sigma: 1 + \omega \rightarrow \omega$. But writing the elements of $1 + \omega$ as

$0' < 0 < 1 < 2 < \cdots$

and the elements of $\omega$ as

$0 < 1 < 2 < 3 < \cdots$,

the isomorphism is clear. We just take $\sigma(0') = 0$ and $\sigma(n) = S(n)$ for all $n \in \omega$.