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Math Help - 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: 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.
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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 \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?
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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: 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...
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  4. #4
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    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}.
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  5. #5
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    Quote Originally Posted by tcRom View Post
    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?
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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...

    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.
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