about a club set I hope someone could help me with my question.
I have to prove that $$C=\{ \alpha < \omega_1 : \alpha \in LIM\text{ and }j\upharpoonright \alpha \times \alpha : \alpha \times \alpha \leftrightarrow \alpha) \}$$ is club in $\omega_1$, where $j : \omega_1 \times \omega_1 \leftrightarrow \omega_1$.
So far what I've done is to prove that $C$ is closed, but I'm not sure that the prove I gave its ok, I don't know how I can prove that is unbounded. I thought that  by contradiction was the easiest way, but I got stuck in the middle of my prove.I need help with this particular thing because I  will have to use it for another demonstration I've been working on.

Let $\alpha<\omega_1$ be given. Find a countable elementary submodel $M\prec H(\theta)$ (for some large enough $\theta$) containing $\alpha$ and  $j$ as elements then it is easy to see that $M\cap\omega_1=\delta\in C$  and $\alpha<\delta$.
If you are not familiar with elementary submodels then you can proceed as follows:
Let $\alpha$ be given. Construct recursively a strictly increasing sequence $\alpha_n (n\in\omega)$   as follows: Choose $\alpha_0\in\omega_1$ so that $j[\alpha\times\alpha]\cup j^{-1}[\alpha]\subseteq \alpha_0$. Suppose that $\alpha_n$ has been defined. Let $\alpha_{n+1}$ be any countable ordinal so that  $j[\alpha_n\times\alpha_n]\subset \alpha_{n+1}$ and $ j^{-1}[\alpha_n]\subseteq \alpha_{n+1}\times \alpha_{n+1}$. It is easy to verify that $\sup_n \alpha_n\in C$ and is bigger than $\alpha$. Thus $C$ is unbounded.
