Doubts about tower number. We say that the sequense $\{T_\alpha\}_{\alpha<\lambda}\subseteq[\omega]^\omega$ (where $\lambda$ is a ordinal) is a tower iff


*

*$\alpha<\beta<\lambda\rightarrow T_\beta\subseteq^*T_\alpha$.

*$\neg\exists K\in[\omega]^\omega\forall\alpha<\lambda(K\subseteq^*T_\alpha)$.


Here "$A\subseteq^*B$" means: $\exists C\in [A]^{<\omega}(A\setminus C\subseteq B)$.
With this, we define the number tower as the least length of a tower. 
My doubt is, how can i to show that the number $\mathfrak{t}$ is well defined?
 A: In order for $\mathfrak{t}$ to be well-defined, you only need to see that a tower exists. Consider, for sequences $(T_{\alpha} \mid \alpha < \lambda)$, $T_\alpha \subseteq [\omega]^{\omega}$, the property
$$
\alpha < \beta < \lambda \implies T_{\beta} \subseteq^* T_{\alpha} \wedge T_\beta \neq T_\alpha.
\tag{$\dagger$}
$$
Since $(\dagger)$ is preserved under unions and any such sequence has length $<(2^{\aleph_0})^{+}$, there is a maximal sequence $(T_{\alpha} \mid \alpha < \lambda)$ satisfying $(\dagger)$. It's easy to see that this sequence is a tower. 
(If there were a counterexample $K$, we could define
$$
T_{\lambda} := \begin{cases}
K & \text{, if } \lambda \text{ is a limit ordinal} \\
K \setminus \min T_{\lambda -1 } & \text{, otherwise}
\end{cases}
$$ and then $(T_{\alpha} \mid \alpha < \lambda + 1)$ would satisfy $(\dagger)$.)
A: You could weaken this even more. Why require that $\lambda$ is an ordinal? Why not ask about just a chain without a lower bound, and not necessarily a well-ordered chain? The answer is that all of this doesn't matter.
This follows from the following observations:


*

*Every linear order has a cofinal well-order.

*The least order type of a cofinal well-order is itself a regular cardinal.


Combine these, and you get that the least order type of a tower is exactly the least cardinal. For your question, of course, the second observation is enough to show that for just well-ordered types, considering the cardinality is enough.
