How can an ordinal have a biunivocal correspondence with an ordinal that precedes it?

Question

if an ordinal a precedes an ordinal b, that means than a is an element in b, and every element in a is also in b. How can there be two different ordinals with the same magnitud? How can an ordinal have a biunivocal correspondence with an ordinal that precedes it? isn’t b, necessary bigger than a?

i’m struggling with the following definition of cardinal “[ordinals that] possess the property of tolerating no one-to-one correspondence with any of the ordinals which precede them”. isn’t that the case for every ordinal?

sorry I’m not actually studying set theory; I must have missed something.

Answer 1

It is not the case for any ordinal.

Consider the ordinal $\omega + 1 = s(\omega) = \mathbb{N} \cup \{\mathbb{N}\}$. Then $\omega+1$ and $\omega = \mathbb{N}$ are in bijection with each other by the bijection

$f(\mathbb{N}) = 0$
$f(n) = s(n)$ for all $n \in \mathbb{N}$

But clearly $\omega$ precedes $\omega + 1$.

Answer 2

Mark Saving gave in his answer the example of a successor ordinal that is not a cardinal. By the way, this is a general fact that an infinite successor ordinal can't be a cardinal.

$\omega + \omega$ is an example of a limit ordinal that is not a cardinal as its cardinality is $\omega$.