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?
Ordered pairs in a power set
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elementary-set-theory
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2Just a quick note, my solution to this particular problem was wrong. I'm going to get around to editing soon. – 2011-07-03
1 Answers
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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 ${\mathcal P}(T)$.