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Math Help - Well-ordered sets.

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Well-ordered sets.

    A question on theme of well-ordered sets.

    Lets say I have a set B that contains the all odd natural numbers.

    Also, a set D\subseteq\mathbb{N}.

    Given that B\cup D \neq \emptyset.

    Let b be a first element in B\cup D in <B,< >.

    We define the order \prec _0 on \mathbb{N} be the set: \{0,2,4,6...,1,3,5,7,...\}


    I need to show that b is first element in D in <\mathbb{N},\prec _0>

    -------------------------------------------------------------------------------

    I started by saying that if c\in D, and c odd, then c\in B\cup D , hence b is first.

    But what I do with the case in which c is even...?


    Thank you for your time.
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  2. #2
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Given that B\cup D \neq \emptyset.
    Do you mean the intersection? Since B is nonempty, the union is known to be nonempty.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    A question on theme of well-ordered sets.

    Lets say I have a set B that contains the all odd natural numbers.

    Also, a set D\subseteq\mathbb{N}.

    Given that B\cup D \neq \emptyset.

    Let b be a first element in B\cup D in <B,< >.

    We define the order \prec _0 on \mathbb{N} be the set: \{0,2,4,6...,1,3,5,7,...\}


    I need to show that b is first element in D in <\mathbb{N},\prec _0>

    -------------------------------------------------------------------------------

    I started by saying that if c\in D, and c odd, then c\in B\cup D , hence b is first.

    But what I do with the case in which c is even...?


    Thank you for your time.

    I did a mistake on the original post of mine, not surprising fact...

    Lets say I have a set B that contains the all odd natural numbers.

    Also, a set D\subseteq\mathbb{N}.

    Given that B\cap D \neq \emptyset.

    Let b be a first element in B\cap D in <B,< >.

    We define the order \prec _0 on \mathbb{N} be the set: \{0,2,4,6...,1,3,5,7,...\}


    I need to show that b is first element in D in <\mathbb{N},\prec _0>

    -------------------------------------------------------------------------------

    I started by saying that if c\in D, and c odd, then c\in B\cap D , hence b is first.

    But what I do with the case in which c is even...?



    Thanks again!
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  4. #4
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    Suppose that B is just the odds and D=\{0,1,2,3,4,5,6,7,8,9\}.
    What is the first element in B\cap D~?

    What is the first element in D in <\mathbb{N},\prec_0>~?

    What are you trying to prove?
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  5. #5
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Plato View Post
    Suppose that B is just the odds and D=\{0,1,2,3,4,5,6,7,8,9\}.
    What is the first element in B\cap D~?

    What is the first element in D in <\mathbb{N},\prec_0>~?

    What are you trying to prove?
    Answers:

    1. The first element in B\cap D is 0.

    2. D in <\mathbb{N},\prec_0> is the set: \{0,2,4,6,8,1,3,5,7,9\}. And 0 is first.

    3. I'm trying to prove that the ordered set <\mathbb{N},\prec_0> is well ordered.


    So what I have done so far:


    Let A be a set of all even numbers. A\subset \mathbb{N}, hence <A,<> well ordered.

    Let B be a set of all odd numbers. B\subset \mathbb{N}, hence <B,<> well ordered.

    Let D be non-empty subset of \mathbb{N}.

    Let we look on A\cap D. Suppose A\cap D \neq \emptyset. Let a be first element in A\cap D, in well ordered set <A,<>.

    a is first element in D in <\mathbb{N},\prec_0>.

    Explanation for my last sentence:

    For all b element of D which is different from a, if b is even, then b\in A\cap D, hence a is first.
    If b is odd, a is first by definition of order \prec_0.

    Now, if A\cap D = \emptyset then B\cap D \neq \emptyset.
    Let b be first element in B\cap D in <B,<>.

    How can I show that b is first element in D in <\mathbb{N},\prec_0> ?









    Thanks again!
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  6. #6
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Answers:
    1. The first element in B\cap D is 0.
    0 is not an odd number
    0\notin B. How can 0\in B\cap\mathbb{N}~?

    Your notation is completely off base.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Answers:

    1. The first element in B\cap D is 0.

    2. D in <\mathbb{N},\prec_0> is the set: \{0,2,4,6,8,1,3,5,7,9\}. And 0 is first.

    3. I'm trying to prove that the ordered set <\mathbb{N},\prec_0> is well ordered.


    So what I have done so far:


    Let A be a set of all even numbers. A\subset \mathbb{N}, hence <A,<> well ordered.

    Let B be a set of all odd numbers. B\subset \mathbb{N}, hence <B,<> well ordered.

    Let D be non-empty subset of \mathbb{N}.

    Let we look on A\cap D. Suppose A\cap D \neq \emptyset. Let a be first element in A\cap D, in well ordered set <A,<>.

    a is first element in D in <\mathbb{N},\prec_0>.

    Explanation for my last sentence:

    For all b element of D which is different from a, if b is even, then b\in A\cap D, hence a is first.
    If b is odd, a is first by definition of order \prec_0.

    Now, if A\cap D = \emptyset then B\cap D \neq \emptyset.
    Let b be first element in B\cap D in <B,<>.

    How can I show that b is first element in D in <\mathbb{N},\prec_0> ?









    Thanks again!
    This is still very confusing to me. Are you saying that \prec_0 is just the ordering which can be explained 'First 2\mathbb{N} and then 2\mathbb{N}+1"? And, so what you're trying to show is that if one considers D as a subset of \left(D,\prec_0\right) that it has a least element...and moreover that its least element coincides with its least element as a subset of \mathbb{N} with the usual ordering?
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  8. #8
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Plato View Post
    0 is not an odd number
    0\notin B. How can 0\in B\cap\mathbb{N}~?

    Your notation is completely off base.
    You are absolutely right!

    B n D= {1,3,5,7,9} and first element is 1. (1<3<5<7<9)
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  9. #9
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Drexel28 View Post
    This is still very confusing to me. Are you saying that \prec_0 is just the ordering which can be explained 'First 2\mathbb{N} and then 2\mathbb{N}+1"? And, so what you're trying to show is that if one considers D as a subset of \left(D,\prec_0\right) that it has a least element...and moreover that its least element coincides with its least element as a subset of \mathbb{N} with the usual ordering?
    'Yes' for your first question.

    I am trying to show that for any non-empty subset of N has first element under the order <_0 .


    first element:
    a in < A,< > is first element or just first, if to all b in A a<b.
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  10. #10
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    Let \mathcal{E} be the subset of even numbers in \mathbb{N} and \mathcal{O}=\mathbb{N}\setminus\mathcal{E}
    Now the order in < \mathbb{N},\prec_0> works this way: any two even or odd numbers are ordered in the natural way.
    But any even number precedes every odd number.

    Now consider any nonempty subset D in < \mathbb{N},\prec_0>.
    If D\cap\mathcal{E}\not=\emptyset then the first even in D is its first element.
    Else wise, the first odd number in D is its first element.
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  11. #11
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Plato View Post
    Let \mathcal{E} be the subset of even numbers in \mathbb{N} and \mathcal{O}=\mathbb{N}\setminus\mathcal{E}
    Now the order in < \mathbb{N},\prec_0> works this way: any two even or odd numbers are ordered in the natural way.
    But any even number precedes every odd number.

    Now consider any nonempty subset D in < \mathbb{N},\prec_0>.
    If D\cap\mathcal{E}\not=\emptyset then the first even in D is its first element.
    Else wise, the first odd number in D is its first element.
    Thank you very much Plato!
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