What is $\aleph_1$ in the context of this definition

Question

I would appreciate help understanding what $\aleph_1$ is according to this definition:

If $\alpha$ is an ordinal, then $\aleph_{\alpha}$ is the unique infinite cardinal such that: $\{\kappa:\kappa\text{ is an infinite cardinal and }\kappa\lt\aleph_{\alpha}\}$ is isomorphic to $\alpha$ as a well-ordered set.

My question specifically is:

With $\alpha=1$ an ordinal ($=\{0\}$ according to von Neumann), what would the set $\{\kappa:\kappa\text{ is an infinite cardinal and }\kappa\lt\aleph_1\}$ look like?

EDIT I think I should emphasize that the aspect that especially confuses me is "isomorphic to $\alpha=1$."

Thanks

Answer 1

Note that $1$ is just $\{0\}$. So a well-ordered set is isomorphic to $1$ if and only if it has exactly one element.

In the case of $\aleph_1$, it is the unique transfinite cardinal $\kappa$ which has exactly one transfinite cardinal smaller than it. In other words, it would be exactly the smallest cardinal which is larger than $\aleph_0$.

Answer 2

As $\aleph_1$ is the second infinite cardinal, the infinite cardinals $\le\aleph_1$ are $\aleph_0$ and $\aleph_1$. It looks like your definition should have $<$ rather than $\le$.