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Math Help - Maximal Element

  1. #1
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    Maximal Element

    Fix N, some integer, and suppose L is a nonempty set of integers such that every a in L is <N. Show that L has a maximal element.

    L = \{n \in N|n\geq a \ \forall a\in L\}

    How do I show this?
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    Fix N, some integer, and suppose L is a nonempty set of integers such that every a in L is <N. Show that L has a maximal element.
    L = \{n \in N|n\geq a \ \forall a\in L\}
    I think you need to edit the OP.
    The set L you have given does not meet the description: "Fix N, some integer, and suppose L is a nonempty set of integers such that every a in L is <N"
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  3. #3
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    That is verbatim from the book.
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  4. #4
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    I am assuming the question goes like this.
    If N\in\mathbb{Z} then define \mathcal{L}=\{n\in\mathbb{Z}:n<N\}.
    You can surely see that N\notin\mathcal{L}\text{ and }(N-1)\in\mathcal{L}. HOW & WHY?
    What can you say about N-1~?
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  5. #5
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    Quote Originally Posted by Plato View Post
    I am assuming the question goes like this.
    If N\in\mathbb{Z} then define \mathcal{L}=\{n\in\mathbb{Z}:n<N\}.
    You can surely see that N\notin\mathcal{L}\text{ and }(N-1)\in\mathcal{L}. HOW & WHY?
    What can you say about N-1~?
    n\leq N-1<N

    N-1 can be greater than n or equal to it.
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  6. #6
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    i am a bit confused. the set L = {n in Z: n < N} is the set of ALL integers less than N, whereas in the orginial question, L is presumed to be merely some (non-emtpy) set with the property that:

    a in L--> a < N, which is NOT the same thing.

    in other words, we are only talking about some subset of {n in Z: n < N}, and we don't know WHICH subset.

    i would think it would be easier to assume our set L has no maximal element (that is, if a is in L, then there is b in L with a < b

    note that a+1 ≤ b, necessarily), and from this assumption, show that some member in L must be larger than N,

    which is a contradiction, so L must have a maximal element.
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