# Well-ordered sets.

• Mar 19th 2011, 04:57 PM
Also sprach Zarathustra
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 $$.

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

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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...?

• Mar 20th 2011, 12:30 PM
emakarov
Quote:

Originally Posted by Also sprach Zarathustra
Given that $B\cup D \neq \emptyset$.

Do you mean the intersection? Since B is nonempty, the union is known to be nonempty.
• Mar 21st 2011, 09:52 AM
Also sprach Zarathustra
Quote:

Originally Posted by Also sprach Zarathustra
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 $$.

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...?

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 $$.

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!
• Mar 21st 2011, 10:16 AM
Plato
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?
• Mar 21st 2011, 12:12 PM
Also sprach Zarathustra
Quote:

Originally Posted by Plato
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?

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 $$ well ordered.

Let $B$ be a set of all odd numbers. $B\subset \mathbb{N}$, hence $$ 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 $$.

$a$ 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 $$.

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

Thanks again!
• Mar 21st 2011, 12:20 PM
Plato
Quote:

Originally Posted by Also sprach Zarathustra
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.
• Mar 21st 2011, 12:22 PM
Drexel28
Quote:

Originally Posted by Also sprach Zarathustra

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 $$ well ordered.

Let $B$ be a set of all odd numbers. $B\subset \mathbb{N}$, hence $$ 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 $$.

$a$ 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 $$.

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?
• Mar 21st 2011, 01:37 PM
Also sprach Zarathustra
Quote:

Originally Posted by Plato
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)
• Mar 21st 2011, 01:45 PM
Also sprach Zarathustra
Quote:

Originally Posted by Drexel28
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?

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.
• Mar 21st 2011, 02:36 PM
Plato
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.
• Mar 21st 2011, 03:46 PM
Also sprach Zarathustra
Quote:

Originally Posted by Plato
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!