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Let $A$ be a countable set equipped with a partial order $\prec$. We all know that there are subsets $C\subset A$ with the property that the order $\prec$ in $C$ is total. In other words, any two elements $a$ and $b$ in $C$ are comparable.

Let $\mathscr{C}$ be the collection of all subsets $C$ of $A$ that are totally ordered.

Question 1: Is it true $A=\displaystyle\bigcup_{C\in\mathscr{C}}C$?

Denote by $\mathscr{D}$ a any subcolection $\mathscr{D}\subset \mathscr{C}$ such that for all $C_1,C_2\in\mathscr{D}$ we have empty intersection $C_1\cap C_2=\emptyset$.

Question 2: There is a subcollection $\mathscr{D}$ as described above such that $A=\displaystyle\bigcup_{C\in\mathscr{D}}C$?

I know these questions seem intuitive but I'd be happy with some demonstration of elementary set theory either by reduction to absurdity or not

Thank's.

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1 Answer 1

up vote 3 down vote accepted

They answer to the first question is trivially yes, since every singleton is linearly ordered.

Revised to match edited question:

The answer to the second question is also trivially yes: let $\mathscr{D}=\{\{a\}:a\in A\}$.

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If $\mathscr{D}$ consist of just one singleton were is the contradiction? Can you give an another counter example? –  Elias Nov 9 '12 at 21:36
    
@Elias: Suppose that $A=\{0,1\}$ with either of the two possible partial orders, and $\mathscr{D}=\{\{0\}\}$. Clearly $A\ne\biguplus_{C\in\mathscr{D}}$. –  Brian M. Scott Nov 9 '12 at 21:40
    
Sorry it was a mistake of mine. I've edited my question. Thank's. –  Elias Nov 9 '12 at 21:49
    
@Elias: I’ve updated my answer to match. –  Brian M. Scott Nov 9 '12 at 21:54
1  
@Elias: It’s not always true if you make that further assumption. Remember, the relation of equality is a partial order, so it may be that the only subsets of $A$ that are linearly ordered are the singletons. –  Brian M. Scott Nov 9 '12 at 22:18

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