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Thread: Show S is closed and unbounded in omega 1

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

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

    If S were not closed, then $\displaystyle \exists A \in S|sup(A)> S$. 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 $\displaystyle \exists \alpha \in \omega_{1}| \forall \beta \in S, \alpha > \beta$. Choose $\displaystyle \beta \in S|sup(\beta)=\omega_{1}$, then there does not exist $\displaystyle \alpha \in \omega_{1}| \alpha > \beta$, since $\displaystyle \alpha = \beta$ 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 $\displaystyle \beta \in \omega_{1}$ that has $\displaystyle sup(\beta)=\omega_{1}$. So, I need help showing S is unbounded for this possibility. Any takers on helping? Let me know if something is unclear.
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  2. #2
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    Let's see if someone can find any holes in the following:

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

    Therefore, S must be closed and unbounded. Da?
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  3. #3
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    It was a long time ago that I did set theory...

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

    If S were not closed, then $\displaystyle \exists A \in S|sup(A)> S$. 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 $\displaystyle A\in S$" or "$\displaystyle A\subseteq S$"? Not every subset is a member, is it?

    Finally, your proof does not seem to use $\displaystyle f$ at all...
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  4. #4
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    Yes, $\displaystyle f$ is an arbitrary function and yes, $\displaystyle f(S) \subseteq S $.

    We are basically showing that if $\displaystyle \alpha$ is closed under $\displaystyle f$, and the set S is made from these $\displaystyle \alpha$'s, then S is also closed. So in a sense, we are extending closure from the $\displaystyle \alpha$'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 $\displaystyle A \subseteq S$, then $\displaystyle sup(A) \in S$ or $\displaystyle sup(A) = \omega_{1}$.
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  5. #5
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    Quote Originally Posted by tcRom View Post
    We are basically showing that if $\displaystyle \alpha$ is closed under $\displaystyle f$, and the set S is made from these $\displaystyle \alpha$'s, then S is also closed.
    But the word closed is used in two different senses here, isn't it? $\displaystyle \alpha$ is closed under $\displaystyle f$, i.e., $\displaystyle f(\alpha)\subseteq\alpha$, whereas $\displaystyle S$ 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 $\displaystyle f:\omega_1\to\omega_1$, $\displaystyle f$ produces ordinals, not just arbitrary sets. On ordinals, $\displaystyle \alpha\subseteq\beta$ means $\displaystyle \alpha\in\beta$ or $\displaystyle \alpha=\beta$, i.e., $\displaystyle \alpha \le\beta$. Now consider $\displaystyle f$ such that $\displaystyle f(\alpha)=0$ for all $\displaystyle \alpha\in\omega$ and $\displaystyle f(\beta)=\beta+1$ for $\displaystyle \beta\notin\omega$. Then $\displaystyle f(\alpha)\le\alpha$ on $\displaystyle \omega$, so $\displaystyle \omega\subseteq S$. However, $\displaystyle f(\sup\{\alpha\mid\alpha\in\omega\})=f(\omega)=\om ega+1$, so $\displaystyle f(\sup\omega)\not\le\sup\omega$, and so $\displaystyle \omega\notin S$. How can then $\displaystyle S$ be closed?
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  6. #6
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    I got more clarification from my professor because now I'm rather confused about closed...

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

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

    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:$\displaystyle \forall\beta\in\omega_{1}, \exists\alpha\in\omega_{1}, \alpha<\beta, \alpha\in S$

    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.
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