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On page 36 of his Set Theory book, Halmos states the following (with $X$ and $Y$ being two sets whose cartesian product we'll consider):

"Consider any particular unordered pair {a, b} with $a\neq b$, and consider the set Z of all families z, indexed by {a, b}, such that $z_a \in X$ and $z_b \in Y$. If the function $f$ from $Z$ to $X \times Y$ is defined by $f(z) = (z_a, z_b)$, then $f$ is a one-to-one correspondence between $X \times Y$ and $Z$."

My question is: where is this one-to-one correspondence? The way I'm interpreting this is the following:

We have a set of families $Z$, which is the range of some family {$ z_i $}, and where each term of this family ($Z$) is also a family. Then, the domain of such family {$ z_i $} is the index $I=\, ${$a, b$}; such that the terms of the family will be $z_a$ and $z_b$. Then, if $z_a \in X$ and $z_b \in Y$, if we define a function $f: Z\longrightarrow X\times Y$, where $f(z):= (z_a, z_b)$, then, whatever $z$ we choose from $Z$, will its image be the fixed ordered pair $(z_a, z_b)$?

Sorry if I'm overcomplicating this, I just started studying families and this has me a bit confused.

Thanks in advance.

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A family of elements indexed by the two element set $\{a,b\}$ is just a function with domain $\{a,b\}$, and for such a function $z$, the index notation simply means $z_a=z(a),\ z_b=z(b)$. So the set $Z$ is $$Z=\{z:\{a,b\}\to X\cup Y: z(a)\in X,\,z(b)\in Y\}\,.$$ The correspondence to $X\times Y$ is given by $z\mapsto (z(a),\,z(b))$ and $(x,y)\mapsto (a\mapsto x,\ b\mapsto y)$.

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  • $\begingroup$ Thanks! I thought I had it, but I'm confused: what does the second part of the correspondence mean? Namely the part where (x, y) is an element of the domain, and it's image is (a -> x, b -> y)? (I apologize, I don't know how to TeX in comments) $\endgroup$ – GabrielAguilar.0 Dec 31 '20 at 22:24
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    $\begingroup$ The first map goes $Z\to X\times Y$, the other one is its inverse $X\times Y\to Z$, and $\mapsto$ denotes assignment. $\endgroup$ – Berci Dec 31 '20 at 23:15
  • $\begingroup$ Got it, thanks so much! $\endgroup$ – GabrielAguilar.0 Dec 31 '20 at 23:56

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