# Thread: Show S is closed and unbounded in omega 1

1. ## Show S is closed and unbounded in omega 1

Given: $f:\omega_{1} \rightarrow \omega_{1}, S=\{\alpha \in \omega_{1}: \alpha$ is closed under $f\}$.
Show S is closed and unbounded in $\omega_{1}$.

If S were not closed, then $\exists A \in S|sup(A)> S$. Thus, S must be closed since sup(A) must be in S or equal to S (by the definition of closed).

If S were bounded, then $\exists \alpha \in \omega_{1}| \forall \beta \in S, \alpha > \beta$. Choose $\beta \in S|sup(\beta)=\omega_{1}$, then there does not exist $\alpha \in \omega_{1}| \alpha > \beta$, since $\alpha = \beta$ in this case. Thus, a contradiction and S must be unbounded. Therefore, S is closed and unbounded.

NOW, the problem... The second part dealing with bounded does not cover the case where there is no $\beta \in \omega_{1}$ that has $sup(\beta)=\omega_{1}$. So, I need help showing S is unbounded for this possibility. Any takers on helping? Let me know if something is unclear.

2. Let's see if someone can find any holes in the following:

Let S be bounded in $\omega_{1}$ and let $f:\alpha \rightarrow \alpha, \alpha \in \omega_{1}$ such that $f(\alpha) = 1 + \alpha$. Since $\alpha$ must be closed, as above, under $f$, S is, at the least, comprised of all limit ordinals in $\omega_{1}$. But, there are an unbounded number of limit ordinals in $\omega_{1}$, making S unbounded, a contradiction.

Therefore, S must be closed and unbounded. Da?

3. It was a long time ago that I did set theory...

Originally Posted by tcRom
Given: $f:\omega_{1} \rightarrow \omega_{1}, S=\{\alpha \in \omega_{1}: \alpha$ is closed under $f\}$.
So $f$ is an arbitrary function? And does " $S$ closed under $f$" mean $f[S]\subseteq S$?

If S were not closed, then $\exists A \in S|sup(A)> S$. Thus, S must be closed since sup(A) must be in S or equal to S (by the definition of closed).
It looks to me that you are proving that S is closed from the assumption that S is closed. Am I missing something?

Also, what is the definition of closed? Does it say "for every $A\in S$" or " $A\subseteq S$"? Not every subset is a member, is it?

Finally, your proof does not seem to use $f$ at all...

4. Yes, $f$ is an arbitrary function and yes, $f(S) \subseteq S$.

We are basically showing that if $\alpha$ is closed under $f$, and the set S is made from these $\alpha$'s, then S is also closed. So in a sense, we are extending closure from the $\alpha$'s to S. So yes, we are pretty much using closure to show closure, but more extending it than defining it.

For the definition of closed, I am using: If $A \subseteq S$, then $sup(A) \in S$ or $sup(A) = \omega_{1}$.

5. Originally Posted by tcRom
We are basically showing that if $\alpha$ is closed under $f$, and the set S is made from these $\alpha$'s, then S is also closed.
But the word closed is used in two different senses here, isn't it? $\alpha$ is closed under $f$, i.e., $f(\alpha)\subseteq\alpha$, whereas $S$ is just closed, i.e., if , then or , as you write. It's somewhat confusing to say "S is also closed".

I don't have enough intuition about this. E.g., why is the following not a counterexample? Since $f:\omega_1\to\omega_1$, $f$ produces ordinals, not just arbitrary sets. On ordinals, $\alpha\subseteq\beta$ means $\alpha\in\beta$ or $\alpha=\beta$, i.e., $\alpha \le\beta$. Now consider $f$ such that $f(\alpha)=0$ for all $\alpha\in\omega$ and $f(\beta)=\beta+1$ for $\beta\notin\omega$. Then $f(\alpha)\le\alpha$ on $\omega$, so $\omega\subseteq S$. However, $f(\sup\{\alpha\mid\alpha\in\omega\})=f(\omega)=\om ega+1$, so $f(\sup\omega)\not\le\sup\omega$, and so $\omega\notin S$. How can then $S$ be closed?

6. I got more clarification from my professor because now I'm rather confused about closed...

$f:\omega_{1} \rightarrow \omega_{1}, S=\{\alpha \in \omega_{1}: \alpha$ is closed under $f\}$

Means:
1) $f$ restricted to $\alpha: \alpha \rightarrow \alpha$
2) $\forall\beta<\alpha(f(\beta)<\alpha)$

And I have an additional note above S that says "inputs to f below alpha, closed under alpha."

Then, the following definitions are to be used:
Closed:If , then or .
Unbounded: $\forall\beta\in\omega_{1}, \exists\alpha\in\omega_{1}, \alpha<\beta, \alpha\in S$

That's really all I'm given and I'm sorry that I am unable to explain it better or provide better clarification. I'll do my best though if there are further questions.