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The exercise 48 of Chapter 3 in Enderton's Axiomatic Set Theory says: Let $T$ be the set $$\{\varnothing, \{\varnothing\}\}.$$ Find all of the ordered pairs, if any, in $$\mathcal{P}T=\{\varnothing, \{\varnothing\}, \{\{\varnothing\}\}, T\}.$$ The set $T$ does not seem anything like the Kuratowski's definition. I think Enderton is trying to say "There is an ordered pair." I just can't see how there could be an ordered pair since $\mathcal{P}T$ is not a function or relation. Could you please help me see?

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The book is probably titled Elements of Set Theory. –  lhf Jun 28 '11 at 22:46
    
If you are going through Enderton's book, this file might be of interest: docs.google.com/… It was wrttien by yunone math.stackexchange.com/users/1583/yunone –  Martin Sleziak Jul 3 '11 at 12:14
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Just a quick note, my solution to this particular problem was wrong. I'm going to get around to editing soon. –  yunone Jul 3 '11 at 19:35

1 Answer 1

up vote 9 down vote accepted

The set-theoretic ordered pair $(a,b)$ is $\{\{a\},\{a,b\}\}$. Note that this is $$\{\{a\},\{a,a\}\}=\{\{a\},\{a\}\}=\{\{a\}\}$$ if $a=b$. So you have at least one pair in ${\mathcal P}(T)$, namely $(\emptyset,\emptyset)=\{\{\emptyset\}\}$. Since the empty set is not an ordered pair, and no element of an ordered pair is empty, this is the only ordered pair in ${\mathcal P}(T)$.

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