Prove that set $C=\{\alpha<\omega_1:f|\alpha:\alpha\xrightarrow{\rm 1:1,onto}\alpha\times\alpha\}$ is closed and unbounded. Let $f:\omega_1\to\omega_1\times\omega_1$ be bijection.
Prove that set: $$C=\{\alpha<\omega_1:f|\alpha:\alpha\xrightarrow{\rm 1:1,onto}\alpha\times\alpha\}$$
is closed and unbounded.
I do not have any clear idea how to solve such kind of problems, so I would be thankful if someone could explain it to me.
My idea wast first to check if this set is unbounded. So I assumed that there is no $\alpha>0$ such that $\alpha\in C$. So I assumed that for every $\alpha$ we have $\beta_\alpha<\alpha$ such that $f|\alpha:\alpha\to\alpha\times\beta_\alpha$ is bijection. I hoped for some contradiction with the fact that $f$ bijection by taking sequence of $\alpha_k$ going to $\omega_1$ but I do not see such thing.
But let's say that I know that $C$ is unbounded. How to check that it is closed? Should it look like this: let $\alpha$ be limit ordinal and $\alpha\in C$ then of course $\alpha=\sup(C\cap\alpha)\in C$.
Lets take any sequence $\alpha_k\in C$ such that $\lim\alpha_k=\alpha$ and $\alpha$ is limit ordinal. Then $\alpha=\sup(C\cap\alpha)$. But we have sequence of bijective functions $f|\alpha_k:\alpha_k\times\alpha_k$ and from the fact that $\lim\alpha_k=\alpha$ we have $f|\alpha:\alpha\to\alpha\times\alpha$ so $\alpha\in C$
 A: This really just the same method you would use for any other set you want to prove it is closed and unbounded (or, a club).
Unbounded:
Fix some ordinal $\alpha_0$. Now define by recursion $\alpha_{n+1}$ as follows:
Let $\zeta_n$ be the least such that $f"\alpha_n\subseteq\zeta_n\times\zeta_n$, and let $\alpha_{n+1}$ be the largest of these two options:

*

*$\sup\{\beta\mid f(\beta)\in(\zeta_n+1)\times(\zeta_n+1)\}$, or

*$\alpha_n+1$.

Now let $\alpha=\sup\alpha_n$, then it is necessarily the case that $\alpha>\alpha_0$, by the second clause of the defintiion, and I claim that $f\restriction\alpha\to\alpha\times\alpha$ is surjective (it is injective by virtue of being a restriction of an injective function).
To see why it is indeed into $\alpha\times\alpha$, suppose that $\beta<\alpha$, then there is some $n$ such that $\beta<\alpha_n$. Now that means that $f(\beta)\in\zeta_n\times\zeta_n$, and $\zeta_n\leq\alpha_{n+1}$, which means $f(\beta)\in\alpha\times\alpha$.
Next, if $\langle\gamma,\delta\rangle\in\alpha\times\alpha$, there is some $n$ such that both $\gamma$ and $\delta$ are smaller than $\alpha_n$. But now by definition, there is some $\beta<\alpha_{n+1}$ such that $f(\beta)=\langle\gamma,\delta\rangle$.
Closed:
Suppose that $\alpha_n\in C$ is an increasing sequence and $\alpha=\sup\alpha_n$. We want to show that $\alpha\in C$.
This one is slightly easier, since we have continuity (with respect to unions) in several aspects here, since $f\restriction\alpha_n$ form an increasing chain of functions, $f\restriction\alpha$ inherits many properties:

*

*The range of $f\restriction\alpha$ is the union of the ranges $f\restriction\alpha_n$.


*Incidentally, $\alpha\times\alpha=\bigcup(\alpha_n\times\alpha_n)$.
I'll leave you to work out the nitty–gritty details.

Conclusions:
This is really just a model theoretic thing, where you simply want an elementary submodel whose domain which is an ordinal. The elementary submodels form a club in $\mathcal P_{\aleph_1}(\omega_1)$, and $\omega_1$ itself (as the set of countable ordinals) is also a club there. So their intersection is a club.
