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?

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 $(\dagger)$.)

• There is a explicit example of a tower with size $2^{\aleph_0}$? – Gödel Dec 14 '17 at 17:27
• Stephan, the sequence cannot have length $(2^{\aleph_0})^+$ purely on cardinality grounds, but it still needs an (admitedly, short) argument to see that the least length is a cardinal and therefore $\le 2^{\aleph_0}$. – Andrés E. Caicedo Dec 14 '17 at 18:01
• @AndrésE.Caicedo Yes, sure. I didn't put any emphasize on this because for the argument all we needed is that there is some upper bound (and I was about to leave to watch the new Star Wars -- barely managed to hit send before I got picked up). – Stefan Mesken Dec 14 '17 at 22:21
• @Stefan My point is that it is incorrect to say that "any such sequence has length $\le2^{\aleph_0}$". (Don't tell me anything about the movie, though :-) I hasn't opened here yet.) – Andrés E. Caicedo Dec 14 '17 at 22:30
• @AndrésE.Caicedo I've corrected the upper bound. – Stefan Mesken Dec 14 '17 at 22:33

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:

1. Every linear order has a cofinal well-order.
2. 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.