Number of distinct order types corresponding to a finite cardinal number. This question is from "Theory of Sets" by E.Kamke. This is about Type classes. The notion of type classes was given in this paragraph.

The definition of type class:

...We denote by $\mathfrak T_m$ the set of distinct order types which belong to a given cardinal number $\mathfrak m$ , and call this set the type class belonging to $\mathfrak m$ , ...

Then in the footnote , it says:

Actually , $|\mathfrak T_m| = 2^m$ ; see Hausdorff [1], p. 455.

So , if $\mathfrak m$ is a finite cardinal , like $3$ , then the type class belonging to $3$ will be  $\mathfrak T_3$. And $|\mathfrak T_3| = 2^3 = 8$.
But then , there is another line before this paragraph which says ,

So this paragraph says that to every finite cardinal number , there corresponds precisely one order type. So , $|\mathfrak T_3| = 1$ ?
At this point I am kind of confused. Can someone tell me what I am missing?
 A: I believe the footnote forgets to mention that $\mathfrak{m}$ is an infinite cardinal number.
What is true, is that $|\mathfrak{T}_\mathfrak{m}|=2^{\mathfrak{m}}$ for infinite cardinal numbers $\mathfrak{m}$. This is what is mentioned in Hausdorff's book:


The proof that the power of $T(\aleph_0)\geq \aleph$ is by Cantor; that it is $\leq\aleph$ is by Bernstein; both are described in Bernsteins dissertation, Untersuchungen aus der Mengenlehre (Halle 1901). The theorem can be generalised without difficulty that $T(\aleph_\alpha)$ has power $2^{\aleph_\alpha}$;

Note that Hausdorff uses $\aleph$ to denote the power of the continuum $2^{\aleph_0}$. In modern usage this is usually called the cardinality of the continuum (to avoid confusion with the power operation), and usually denoted with the symbol $\mathfrak{c}$.
One is also more or less obligated to note that $|\mathfrak{T}_\mathfrak{m}|=2^{\mathfrak{m}}$ requires the Axiom of Choice, since without Choice not every infinite set can be ordered, and thus $\mathfrak{T}_\mathfrak{m}$ may be empty.
