Total Order

[Ethan3141]

Question:
For [MATH]A,B\subseteq\mathbb{R}[/MATH] we define [MATH]A{\le}B:\iff[/MATH] [MATH]A[/MATH] is a subset of [MATH]B[/MATH]
Show that '[MATH]\le[/MATH]' is not a total order on [MATH]\mathcal{P}(\mathbb{R})[/MATH].

My attempt: So I went down the list of things needed for something to be total order: transitivity, anti-symmetry and totality and noticed totality didn't work

Let [MATH]A=\mathbb{Q},~ B=\mathbb{R}{\setminus}\mathbb{Q}[/MATH]Then [MATH]A{\nsubseteq}B~{\land}~B{\nsubseteq}A{\iff}A{\nleq}B~{\land}~B{\nleq}A[/MATH]Thus '[MATH]\le[/MATH]' is not a total order on [MATH]\mathcal{P}(\mathbb{R})[/MATH].

Notation:

• [MATH]\iff[/MATH] is used as equivalence symbol since [MATH]\leftrightarrow[/MATH] doesn't look right (no double line)
• [MATH]\mathcal{P}[/MATH] is used as power set symbol

I was just wondering if this is correct since the question is worth 20 marks and this doesn't seem like 20 marks worth of reasoning. Any help would be appreciated greatly.

Thanks, Ethan

 

[HallsofIvy]

I think you are interpreting the problem differently that I would. You are proving that there exist a such a relation that is not a total order. I would interpret the problem as showing that any such relation is not a total order.

 

[Ethan3141]

I think you are interpreting the problem differently that I would. You are proving that there exist a such a relation that is not a total order. I would interpret the problem as showing that any such relation is not a total order.

Let [MATH]X[/MATH] ([MATH]\mathcal{P}(\mathbb{R})[/MATH] in this case?) be a set A subset [MATH]O{\subseteq}X{\times}X[/MATH] is a total order if:
i) ... antisymmetry definition ...
ii) ... transitivity definition ...
iii)[MATH]{\forall}x,y{\in}X: (x,y){\in}O\lor(y,x){\in}O[/MATH] <--- totality definition
Does (iii) not imply that I just need to show [MATH]{\exists}x,y{\in}X: (x,y){\not\in}O\land(y,x){\not\in}O[/MATH]
Questions:
1) What is X and O in my case
2) Why do I have to show it for any?
I have a feeling that I don't understand the definition to be honest as the notation it is in is rather weird (Mainly the fact that there's a Cartesian product in it).

Thanks for the reply/help.

 

[Dr.Peterson]

Question:
For [MATH]A,B\subseteq\mathbb{R}[/MATH] we define [MATH]A{\le}B:\iff[/MATH] [MATH]A[/MATH] is a subset of [MATH]B[/MATH]
Show that '[MATH]\le[/MATH]' is not a total order on [MATH]\mathcal{P}(\mathbb{R})[/MATH].

My attempt: So I went down the list of things needed for something to be total order: transitivity, anti-symmetry and totality and noticed totality didn't work

Let [MATH]A=\mathbb{Q},~ B=\mathbb{R}{\setminus}\mathbb{Q}[/MATH]Then [MATH]A{\nsubseteq}B~{\land}~B{\nsubseteq}A{\iff}A{\nleq}B~{\land}~B{\nleq}A[/MATH]Thus '[MATH]\le[/MATH]' is not a total order on [MATH]\mathcal{P}(\mathbb{R})[/MATH].
...
I was just wondering if this is correct since the question is worth 20 marks and this doesn't seem like 20 marks worth of reasoning. Any help would be appreciated greatly.

I would agree with you that showing one pair of sets A and B such that neither [MATH]A\subseteq B[/MATH] nor [MATH]B \subseteq A[/MATH] proves that this is not a total order on the power set. Rather, it is a partial order.

The Cartesian product is just part of the definition of a relation.

And the simplicity of the answer is not a problem; it shows that you understand! They are checking your understanding of the definition, which, as you point out, is not trivial.