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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.

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The proof that $C$ is closed is straightforward. Suppose that $\langle\alpha_n:n\in\omega\rangle$ is a strictly increasing sequence in $C$, and let $\alpha=\sup\{\alpha_n:n\in\omega\}$ Then $$j[\alpha\times\alpha]=j\left[\bigcup_{n\in\omega}\alpha_n\times\alpha_n\right]=\bigcup_{n\in\omega}j[\alpha_n\times\alpha_n]=\bigcup_{n\in\omega}\alpha_n=\alpha\;,$$ so $\alpha\in C$, and $C$ is closed.

One of the standard approaches to showing that a set is unbounded works here. Start with any $\alpha_0\in\omega_1$; you want to find an $\alpha\ge\alpha_0$ such that $j[\alpha\times\alpha]=\alpha$. If $j[\alpha_0\times\alpha_0]=\alpha_0$, you’re done, of course, but you can’t count on that If $j[\alpha_0\times\alpha_0]\nsubseteq\alpha_0$, let $$\alpha_1=\min\{\alpha\in\omega_1:\alpha_1>\alpha_0\land\alpha_0\subseteq j[\alpha_0\times\alpha_0]\subseteq\alpha\}\;.$$ Again, if $j[\alpha_1\times\alpha_1]=\alpha_1$, you’re done, but that’s too much to hope for. In fact, we’ll just ignore the possibility and pretend that it doesn’t happen; as you’ll see, there’s no harm in that. For each $n\in\omega$ simply let $$\alpha_{n+1}=\min\{\alpha\in\omega_1:\alpha_{n+1}>\alpha_n\land\alpha_n\subseteq j[\alpha_n\times\alpha_n]\subseteq\alpha\}\;.$$ The result is a strictly increasing sequence $\langle\alpha_n:n\in\omega\rangle$ in $\omega_1$. Let $\alpha$ be the supremum of this sequence. Then $$j[\alpha\times\alpha]=\bigcup_{n\in\omega}j[\alpha_n\times\alpha_n]\subseteq\bigcup_{n\in\omega}\alpha_{n+1}=\alpha=\bigcup_{n\in\omega}\alpha_n\subseteq\bigcup_{n\in\omega}j[\alpha_{n+1}\times\alpha_{n+1}]=j[\alpha\times\alpha]\;,$$ so $\alpha\in C$.

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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.

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@BrianM.Scott Fixed thanks. –  azarel May 22 '12 at 14:27

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