MHB Strict Total Order for Real Numbers

[evinda]

Hello! (Wave)

A relation R on a set A is called strict total order if it is total order and satisfies the trichotomous identity, i.e. if R is:- antireflexive, so (\forall x \in A) \lnot(xRx) or (\forall x \in A) <x,x> \notin R
- antsymmetric, so (\forall x \in A)(\forall y \in A) (xRy \rightarrow \lnot yRx) or (\forall x \in A) (\forall y \in A) (<x,y> \in R \rightarrow <y,x> \notin R)
- transitive, so (\forall x \in A) (\forall y \in A) (\forall z \in A)(x Ry \wedge yRz \rightarrow xRz)
- (\forall x \in A)(\forall y \in A) \rightarrow xRy \lor yRx \lor x=yAccording to my notes, the relation of total order at the real numbers is a relation of strict total order.\langle_{\mathbb{R}}=\{\langle x,y \rangle \in \mathbb{R}^2: \text{ x is smaller that y}\}But, in this case it will never be x=y, right? (Thinking)

 

[Evgeny.Makarov]

- antireflexive, so (\forall x \in A) \lnot(xRx) or (\forall x \in A) <x,x> \notin R

This property is usually called irreflexivity.

- antsymmetric, so (\forall x \in A)(\forall y \in A) (xRy \rightarrow \lnot yRx) or (\forall x \in A) (\forall y \in A) (<x,y> \in R \rightarrow <y,x> \notin R)

Note that antisymmetry is implies by transitivity and irreflexivity.

According to my notes, the relation of total order at the real numbers is a relation of strict total order.\langle_{\mathbb{R}}=\{\langle x,y \rangle \in \mathbb{R}^2: \text{ x is smaller that y}\}But, in this case it will never be x=y, right?

What will never be x=y? What precisely are you saying and why?

 

[evinda]

This property is usually called irreflexivity.
Note that antisymmetry is implies by transitivity and irreflexivity.

So, if we want to show that a relation is strict total order, does it suffice to show that it is transitive, irreflexive and trichotomous? (Thinking)

What will never be x=y? What precisely are you saying and why?

I meant that if $\langle x,y \rangle \in \langle_{\mathbb{R}}$ it will hold that $x<y$ but it will never hold $x=y$ or $y<x$, right?
So from the identity $(\forall x \in A) (\forall y \in A) \rightarrow xRy \lor yRx \lor x=y$ only this relation will hold: $xRy$. Or am I wrong? (Thinking)

 

[Evgeny.Makarov]

So, if we want to show that a relation is strict total order, does it suffice to show that it is transitive, irreflexive and trichotomous?

Yes.

I meant that if $\langle x,y \rangle \in \langle_{\mathbb{R}}$ it will hold that $x<y$ but it will never hold $x=y$ or $y<x$, right?

Yes.

So from the identity $(\forall x \in A) (\forall y \in A) \rightarrow xRy \lor yRx \lor x=y$ only this relation will hold: $xRy$.

No, that's not correct. Are you claiming that no two numbers are equal?

 

[evinda]

No, that's not correct. Are you claiming that no two numbers are equal?

I thought so, because of the relation we are looking at:

$$\langle_{\mathbb{R}}=\{\langle x,y \rangle \in \mathbb{R}^2: \text{ x is smaller that y}\}$$

If $\langle x,y \rangle \in \langle_{\mathbb{R}}$ it means that $x<y$, right?

That's what I thought that we cannot find $\langle x,y \rangle \in \langle_{\mathbb{R}}$ such that $x=y$ or $y>x$, i.e. $yRx$.

Am I wrong? (Thinking)

 

[Evgeny.Makarov]

If $\langle x,y \rangle \in \langle_{\mathbb{R}}$ it means that $x<y$, right?

Yes.

That's what I thought that we cannot find $\langle x,y \rangle \in \langle_{\mathbb{R}}$ such that $x=y$ or $y>x$, i.e. $yRx$.

Correct. But this does not imply that $\forall x,y\in\Bbb R\;x<y$.

 

[evinda]

Yes.

Correct. But this does not imply that $\forall x,y\in\Bbb R\;x<y$.

I meant $\forall x,y \in R; x<y$, where $R=\langle_{\mathbb{R}}$.
This is right, or not? (Thinking)

 

[Evgeny.Makarov]

I meant $\forall x,y \in R; x<y$, where $R=\langle_{\mathbb{R}}$.
This is right, or not?

This is a tautology (in the dictionary sense). But in post #3, you seemed to make a different claim.

So from the identity $(\forall x \in A) (\forall y \in A) \rightarrow xRy \lor yRx \lor x=y$ only this relation will hold: $xRy$.

Here you seem to be saying that for any $x$ and $y$, $x<y\lor y<x\lor x=y$ holds specifically because the first disjunct, namely, $x<y$ holds. This is wrong.

 

[evinda]

This is a tautology (in the dictionary sense). But in post #3, you seemed to make a different claim.

Here you seem to be saying that for any $x$ and $y$, $x<y\lor y<x\lor x=y$ holds specifically because the first disjunct, namely, $x<y$ holds. This is wrong.

Oh yes, right.. (Nod) So, it will hold $xRy$ or $yRx$ but never $x=y$ where $R=\langle_{\mathbb{R}}$, right? (Thinking)

 

[Evgeny.Makarov]

Are you claiming that no two numbers are equal?

...

 

[evinda]

...

How can it be that $\langle x,y \rangle \in \langle_{\mathbb{R}}$ and $x=y$ ? (Worried)

 

[Evgeny.Makarov]

So, it will hold $xRy$ or $yRx$ but never $x=y$ where $R=\langle_{\mathbb{R}}$, right?

How can it be that $\langle x,y \rangle \in \langle_{\mathbb{R}}$ and $x=y$ ?

You are shifting your claims from one post to the next, which is careless at best. Please write complete formal statements.

Edit: To clarify, by shifting claims I mean a shift from

So from the identity $(\forall x \in A) (\forall y \in A) \rightarrow xRy \lor yRx \lor x=y$ only this relation will hold: $xRy$. Or am I wrong?

in post #3 to

I meant $\forall x,y \in R; x<y$, where $R=\langle_{\mathbb{R}}$.

in post #7. So which is it:
\[
\forall x,y\in\Bbb R\;x<y\lor y<x\lor x=y\qquad(1)
\]
or
\[
\forall \langle x,y\rangle\in R\;x<y\qquad(2)
\]
where $R=\{\langle x,y\rangle\mid x<y\}$? And no, it does not follow from (1) that
\[
\forall x,y\in\Bbb R\;x<y.\qquad(3)
\]
And, yes, (2) is a completely different claim from (3).

 

[evinda]

You are shifting your claims from one post to the next, which is careless at best. Please write complete formal statements.

Aaa, I think that I got it now... (Wait)From the trichotomous identity, we get that $(\forall x \in A) (\forall y \in A) \rightarrow x \langle_{\mathbb{R}} y \lor y \langle_{\mathbb{R}} x \lor x=y$.

When we have two real numbers $a,b$, one of the above cases will necessarily hold.
It will be $a>b$ or $a=b$ or $a<b$.
I thought that the third case, $x=y$ is also related to the relation we have, but isn't.. (Shake) Right? (Thinking)

 

[Evgeny.Makarov]

From the trichotomous identity, we get that $(\forall x \in A) (\forall y \in A) \rightarrow x \langle_{\mathbb{R}} y \lor y \langle_{\mathbb{R}} x \lor x=y$.

When we have two real numbers $a,b$, one of the above cases will necessarily hold.
It will be $a>b$ or $a=b$ or $a<b$.

Yes.

I thought that the third case, $x=y$ is also related to the relation we have, but isn't.. Right?

I am not sure what you mean exactly here.

 

[evinda]

Yes.

(Happy)

I am not sure what you mean exactly here.

From the trichotomous identity we have that $(\forall x \in A) (\forall y \in A) \rightarrow xRy \lor yRx \lor x=y$.
I meant that while $xRy$ and $yRx$ depend on the relation $R$ we consider, the third case $x=y$ does not depend on $R$.

 

[Evgeny.Makarov]

From the trichotomous identity we have that $(\forall x \in A) (\forall y \in A) \rightarrow xRy \lor yRx \lor x=y$.
I meant that while $xRy$ and $yRx$ depend on the relation $R$ we consider, the third case $x=y$ does not depend on $R$.

Well, yes, equality is the same in all laws of trichotomy (in the context of orders) while $R$ changes.

 

[evinda]

Well, yes, equality is the same in all laws of trichotomy (in the context of orders) while $R$ changes.

I see.. (Nod)
So, can we only find relations $R$ which are related to equality that satisfy this identity or are there also other trichotomous relations ? (Thinking)

 

[Evgeny.Makarov]

For all (finite, for simplicity) sets $A$ and $B$, $|A|<|B|\lor |B|<|A|\lor|A|=|B|$. Here $|A|=|B|$ is not the same as $A=B$. In general, of course, there are all kinds of situations where exactly one of three options necessarily holds.

 

[evinda]

For all (finite, for simplicity) sets $A$ and $B$, $|A|<|B|\lor |B|<|A|\lor|A|=|B|$. Here $|A|=|B|$ is not the same as $A=B$. In general, of course, there are all kinds of situations where exactly one of three options necessarily holds.

Could you give me an example of an other relation for which the trichotomy holds? (Thinking)

 

[Evgeny.Makarov]

I suggest you formulate this question precisely enough so that it would be suitable to start a new thread. Please also provide your attempt at solution, and it would be nice to include a rationale for such question.

 

[evinda]

I suggest you formulate this question precisely enough so that it would be suitable to start a new thread. Please also provide your attempt at solution, and it would be nice to include a rationale for such question.

I don't really have an attempt.. (Worried)
I was just thinking about whether it is possible that there is a relation that is not related to an equality or an inequality but satisfies trichotomy since one of its cases is $x=y$ or not... (Thinking)

 

[Evgeny.Makarov]

I was just thinking about whether it is possible that there is a relation that is not related to an equality or an inequality but satisfies trichotomy since one of its cases is $x=y$ or not...

At least state your problem precisely. I am not sure about the meaning of the following phrases:

• "relation related to an equality",
• "relation satisfies trichotomy",
• "the cases of the relation".

 

[evinda]

At least state your problem precisely. I am not sure about the meaning of the following phrases:

• "relation related to an equality",
• "relation satisfies trichotomy",
• "the cases of the relation".

We were looking at the relation:

$$\langle_{\mathbb{R}}=\{\langle x,y \rangle \in \mathbb{R}^2: \text{ x is smaller that y}\}$$

We can easily see that the property would also hold if we would have the relation:

$$\langle_{\mathbb{R}}=\{\langle x,y \rangle \in \mathbb{R}^2: \text{ x is greater that y}\}$$

Both of the above relations have to do with an inequality.

I was wondering if there is an other relation $R$, a one that is not related to $<$ or $>$ but satisfies the property:

$$(\forall x \in A) (\forall y \in A) (xRy \lor x=y \lor yRx)$$