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Show that for all class $ \{A_i \}_{i\in I}$ of $A $, the relation $T$ is equivalence relation where $T$ is defined to be: $xTy$ iff exist $i\in I$ such that $\{ x,y\}\subseteq A_i$.

My attempt:

Reflexivity: $xTx$ because $A_i$ is a class of $A$, and $A_i$ is not empty.
Symmetry: If $xTy$, the there exist $i\in I$ such that $\{ x,y\}\subseteq A_i$. The same $i\in I$ is suitable to $yTx$. Transitivity: Let $xTy$ and $yTz$, so exist $i$ such that $\{ x,y\}\subseteq A_i$ and exist $j$ such that $\{ y,z\}\subseteq A_j$. Now clearly( I believe) $\{ x,z\}\subseteq A_i\Delta A_j$. Now what?

edit: all class $ \{A_i \}_{i\in I}$ of $A $ = All classes of $A$ under the relation $T$. Thank you.

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What does "all class $\{A_i\}_{i\in I}$ of $A$" mean? –  Asaf Karagila Nov 6 '12 at 9:11
    
Are you perhaps looking for something like Proof of a Proposition on Partitions and Equivalence Classes and [Relation induced by partition] at ProofWiki? –  Martin Sleziak Nov 6 '12 at 11:01

1 Answer 1

up vote 1 down vote accepted

I assume that you have a set $\{A_i\}$ of subsets of $A$ such that $\bigcup_i A_i = A$ and $A_i \cap A_j = \emptyset$ for $i \neq j$.

If $\{x, y\} \subset A_i$ and $\{y, z\} \subset A_j$ then using the last property you can show that $i = j$. By contradiction, if $i \neq j$ then $A_i \cap A_j \supset \{y\} \neq \emptyset$. If $\{A_i\}$ does not hold such property, transitivity need not hold either.

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