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 $\displaystyle <\mathbb{N},\prec _0>$

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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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Originally Posted by Also sprach Zarathustra

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

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Do you mean the intersection? Since B is nonempty, the union is known to be nonempty.

Quote:

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Originally Posted by Also sprach Zarathustra

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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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!

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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Originally Posted by Plato

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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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!

Quote:

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Originally Posted by Also sprach Zarathustra

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

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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.

Quote:

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Originally Posted by Also sprach Zarathustra

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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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?

Quote:

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Originally Posted by Plato

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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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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Originally Posted by Drexel28

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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'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.

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.

Quote:

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Originally Posted by Plato

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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Thank you very much Plato!