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Thread: 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 $\displaystyle B$ that contains the all odd natural numbers.

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

    Given that $\displaystyle B\cup D \neq \emptyset$.

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

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


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

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

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

    But what I do with the case in which $\displaystyle c$ 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 $\displaystyle B\cup D \neq \emptyset$.
    Do you mean the intersection? Since B is nonempty, the union is known to be nonempty.
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    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 $\displaystyle B$ that contains the all odd natural numbers.

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

    Given that $\displaystyle B\cup D \neq \emptyset$.

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

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


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

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

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

    But what I do with the case in which $\displaystyle c$ 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 $\displaystyle B$ that contains the all odd natural numbers.

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

    Given that $\displaystyle B\cap D \neq \emptyset$.

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

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


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

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

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

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



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

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

    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 $\displaystyle B$ is just the odds and $\displaystyle D=\{0,1,2,3,4,5,6,7,8,9\}$.
    What is the first element in $\displaystyle B\cap D~?$

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

    What are you trying to prove?
    Answers:

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

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

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


    So what I have done so far:


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

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

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

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

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

    Explanation for my last sentence:

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

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

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









    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 $\displaystyle B\cap D$ is 0.
    0 is not an odd number
    $\displaystyle 0\notin B$. How can $\displaystyle 0\in B\cap\mathbb{N}~?$

    Your notation is completely off base.
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Answers:

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

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

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


    So what I have done so far:


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

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

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

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

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

    Explanation for my last sentence:

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

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

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









    Thanks again!
    This is still very confusing to me. Are you saying that $\displaystyle \prec_0$ is just the ordering which can be explained 'First $\displaystyle 2\mathbb{N}$ and then $\displaystyle 2\mathbb{N}+1$"? And, so what you're trying to show is that if one considers $\displaystyle D$ as a subset of $\displaystyle \left(D,\prec_0\right)$ that it has a least element...and moreover that its least element coincides with its least element as a subset of $\displaystyle \mathbb{N}$ 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
    $\displaystyle 0\notin B$. How can $\displaystyle 0\in B\cap\mathbb{N}~?$

    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
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    Quote Originally Posted by Drexel28 View Post
    This is still very confusing to me. Are you saying that $\displaystyle \prec_0$ is just the ordering which can be explained 'First $\displaystyle 2\mathbb{N}$ and then $\displaystyle 2\mathbb{N}+1$"? And, so what you're trying to show is that if one considers $\displaystyle D$ as a subset of $\displaystyle \left(D,\prec_0\right)$ that it has a least element...and moreover that its least element coincides with its least element as a subset of $\displaystyle \mathbb{N}$ 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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    Let $\displaystyle \mathcal{E}$ be the subset of even numbers in $\displaystyle \mathbb{N}$ and $\displaystyle \mathcal{O}=\mathbb{N}\setminus\mathcal{E}$
    Now the order in $\displaystyle < \mathbb{N},\prec_0>$ 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 $\displaystyle D$ in $\displaystyle < \mathbb{N},\prec_0>$.
    If $\displaystyle D\cap\mathcal{E}\not=\emptyset$ then the first even in $\displaystyle D$ is its first element.
    Else wise, the first odd number in $\displaystyle D$ 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 $\displaystyle \mathcal{E}$ be the subset of even numbers in $\displaystyle \mathbb{N}$ and $\displaystyle \mathcal{O}=\mathbb{N}\setminus\mathcal{E}$
    Now the order in $\displaystyle < \mathbb{N},\prec_0>$ 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 $\displaystyle D$ in $\displaystyle < \mathbb{N},\prec_0>$.
    If $\displaystyle D\cap\mathcal{E}\not=\emptyset$ then the first even in $\displaystyle D$ is its first element.
    Else wise, the first odd number in $\displaystyle D$ is its first element.
    Thank you very much Plato!
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