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Math Help - Is this enough to solve minimal uncountable set problem?

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    Is this enough to solve minimal uncountable set problem?

    Let S_\Omega be the minimal uncountable well-ordered set.

    (a) show that S_\Omega has no largest element.

    Is it enough to say that if there was a largest element in S_\Omega and we were to remove it, we would still be left with an uncountable set and an ordinal set which will then contradict "minimality".

    (b) Show that for every \alpha\in S_\Omega, the subset \{x|\alpha<x\} is uncountable.

    Can I say: Let S=\{x|\alpha<x\} and let T=\{x|x\leqslant\alpha\} be a countable set. Then S_\Omega = S\cup T. Since S_\Omega is uncountable (defined in he beg. of the problem), then either S or T has to be uncountable and seeing as T is already defined as a countable set it follows that S must not be countable. Thus, the set \{x|\alpha<x\} is uncountable.

    Note: I apologize in advance for my latex. The S_Omega is supposed to be S (subscript Cap Omega).
    Last edited by Opalg; September 20th 2010 at 01:21 PM. Reason: fixed LaTeX
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    Quote Originally Posted by tn11631 View Post
    Let S_\Omega be the minimal uncountable well-ordered set.

    (a) show that S_\Omega has no largest element.

    Is it enough to say that if there was a largest element in S_\Omega and we were to remove it, we would still be left with an uncountable set and an ordinal set which will then contradict "minimality".
    Yes, that's fine.

    Quote Originally Posted by tn11631 View Post
    (b) Show that for every \alpha\in S_\Omega, the subset \{x|\alpha<x\} is uncountable.

    Can I say: Let S=\{x|\alpha<x\} and let T=\{x|x\leqslant\alpha\} be a countable set. Then S_\Omega = S\cup T. Since S_\Omega is uncountable (defined in he beg. of the problem), then either S or T has to be uncountable and seeing as T is already defined as a countable set it follows that S must not be countable. Thus, the set \{x|\alpha<x\} is uncountable.
    The way you have written this, it looks as though you are assuming the result that you should be trying to prove. If T is countable then it follows that S is uncountable, as you have correctly shown. But you can't just assume that T is countable, you need to prove it. In fact, that is not hard, because if T was uncountable then S_\Omega would not be minimal.
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