Isn't it a non-strict total order?

[evinda]

Hello! (Wave)

According to my notes:

$\epsilon_{\omega}=\{ \langle m,n \rangle \in \omega^2: m \in n\}$ defines an order on $\omega$ and is symbolized with $<$. The non-strict order of $\omega$ that corresponds to the order $\epsilon_{\omega}$ is symbolized with $\leq$, so:

$$n \leq m \leftrightarrow n \in m \lor n=m\\(n<m \leftrightarrow n \in m)$$

$R$ order on $A$ induces a non-strict order: $R \cup I_A$
$R$ non-strict order on $A$ induces order $R \setminus I_A$​

Proposition:

• For any natural numbers $m,n$: $m \leq n \leftrightarrow m \subset n$.
• For any natural numbers $m,n$: $m<n \leftrightarrow m \subsetneq n(\leftrightarrow m \in n)$

The relation of non-strict order of the natural numbers is identified with the relation of inclusion $\subset_{\omega}$.Therefore we have the following proposition:

The relation $\subset_{\omega}$ is a total order on $\omega$.
We know that for any natural number $n,m,k \in \omega$ it holds:

1. $n \subset n$
2. $n \subset m \wedge m \subset n \rightarrow m=n$
3. $n \subset m \wedge m \subset k \rightarrow n \subset k$ because of transitivity of $n$
4. $\forall n \in \omega: n \subset m \lor m \subset m \lor m=n$

So isn't the relation $\subset_{\omega}$ a non-strict total order on $\omega$ ? Or am I wrong? (Thinking)

 

[Evgeny.Makarov]

$n \subset m \wedge m \subset k \rightarrow n \subset k$ because of transitivity of $n$

Transitivity of what?

So isn't the relation $\subset_{\omega}$ a non-strict total order on $\omega$ ?

Yes.

 

[evinda]

Transitivity of what?
.

Transitivity of $k$, right? (Blush)

 

[Evgeny.Makarov]

How do you use the transitivity of $k$ to show $n \subset m \wedge m \subset k \rightarrow n \subset k$? Can't it be shown in a simpler way?

 

[evinda]

How do you use the transitivity of $k$ to show $n \subset m \wedge m \subset k \rightarrow n \subset k$? Can't it be shown in a simpler way?

We know that $k \in \omega$, so $k$ is transitive. That means that the element of its elements are elements of $k$.
So: if $x \in k$ and $y \in x$ then $y \in k$.
$$n \subset m\leftrightarrow n \in m \lor n=m$$

$$m \subset k \leftrightarrow m \in k \lor m=k$$

If $n=m$ or $m=k$ then it is trivial.

If $n \in m$ and $m \in k$ we have that $n \in k $ that implies that $n \subset k$.

But.. if we have three sets $A,B,C$ with $A \subset B \wedge B \subset C$ then $A \subset C$.
So we don't have to use the transitivity, right? (Thinking)

 

[Evgeny.Makarov]

But.. if we have three sets $A,B,C$ with $A \subset B \wedge B \subset C$ then $A \subset C$.
So we don't have to use the transitivity, right?

Of course.

 

[evinda]

Great! (Happy) Thanks a lot!