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This is an exercise from Devlin, Joy of sets.

Exercise The ordered pair operation $(a,b)$ defines a binary function on sets. The inverse functions to the function are defined as follows $$ \textrm{if } w=(a,b), \textrm{ then } (w)_0=a\ \textrm{ and}\ \ (w)_1=b$$ Prove that if $w$ is an ordered pair, then:

(a) $(w)_0=\bigcup\bigcap w$;

(b) \begin{equation*} (w)_1 =\left\{ \begin{array}{rl} \bigcup \left[\bigcup w-\bigcap w\right] & \text{if } \bigcup w\neq\bigcap w,\\ \bigcup\bigcup w & \text{if } \bigcup w=\bigcap w \end{array} \right. \end{equation*}

Note 1 if $w$ is a set of sets, then $\bigcup w=\{a|(\exists y\in w) (a\in y)\}$

Note2 the author considers an ordered pair as a set, namely $(a,b)=\{\{a\},\{a,b\}\}$

Note3 please don't ask me what I did or where precisely I got lost, since what I should do and why a mathematician should always try to turn trivialities into complicated stuff is a part of this question.

EDIT What I can't understand is the presence of union symbol both in (a) and (b). I mean, if $x$ is a set, then $\bigcup x=x$. So, in our case, $\bigcap w$ is a set, so that $\bigcup\bigcap w=\bigcap w$, so why $\bigcup$ is needed? Similarly for part (b), $\bigcup\bigcup w=\bigcup w$, so why a double union is needed?

Danae Kissinger
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    Note 3 is so funny :) – Xoff Dec 15 '13 at 11:32
  • So your problem is that you don't understand the question? – Asaf Karagila Dec 15 '13 at 16:31
  • @AsafKaragila yes, that is a part of the problem, i've not a solid formation in set theory, so i essentially don't understand what is going on, i can only see a pair and its components, so why all these unions and intersections?; after a very looooong meditation, i finally convinced myself that i got the general sense, a part for the double union symbol in line two of (b), why two unions and not only one?? – Danae Kissinger Dec 15 '13 at 16:38
  • As sets/set operations are constructed/defined axiomatically, one does not (yet) have many notions like ordered pairs, functions etc. which are used in mathematics in an intuitive manner. All these notions have to be defined from the axioms (pairing, union etc.). The complexity of these definitions and the lost of insight into some extent are probably not something mathematicians like particularly, but rather "side effects" of this formal correctness. Once it's shown that these objects/notions exist the way we want them to be, we don't have to think very much about how they "look like" anyway. – Relatable Harmonics Dec 15 '13 at 19:21

1 Answers1

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The union in this context should be seen as the union of the elements of some set, and it is similar for the intersection. (One could write $"\bigcup x = \bigcup_{y \in x} y"$ to obtain a more common notation.)

Hence if $w=\{\{a\},\{a,b\}\}$, then $\bigcap w = \{a\}$ and $\bigcup \bigcap w = a$. (So taking the union is something like "removing one level of brackets").

It gets a little bit uglier if you want to get the second component: You want to "get" $b$ out of $\bigcup w = \{a,b\}$ by removing $a$ ($\bigcup w \setminus \bigcap w = \{a,b\} \setminus \{a\}$) and then taking the union in order to "remove the brackets" as above. However, this does not work if $a=b$, which the case iff $\bigcup w = \{a,b\} = \{a\} = \bigcap w$. In that situation, you just take $\bigcup \bigcup w = \bigcup \{a, b\} = \bigcup \{a\} = a = b$.

Relatable Harmonics
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