Elementary Set Theory – Definition of an Ordered Pair

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"The ordered pair $(a,b)$ is defined to be the set $\{\{a\},\{a,b\}\}$." ~ Hungerford's Algebra (p.6)

I think this is the first time that i've seen this definition. I've read the wiki page. Is it defined this way, as opposed to a definition relating to functions as in a Cartesian product, because this definition is considered more elementary (or foundational) being that it is related directly to sets?

Also, the definition of an ordered $n$-tuple, according to the wiki page seems vague (perhaps i'm misunderstanding it). For an ordered triple it gives the example:

$$(1,2,3) = \{\{(1,2)\},\{(1,2),3\}\}$$

but how do we know this is not the ordered pair $((1,2),3)$? Or is the difference between $(1,2,3)$ and $((1,2),3)$ considered trivial?

Thirdly, and perhaps unrelated, what does it mean for a natural number to be defined

$$2_{\mathbb{N}} = \{\emptyset,\{\emptyset\} \},$$

and is this also done so that we can define $\mathbb{N}$ in terms of sets?

Best Answer

You may be interested in reading Kuratowski's "Set Theory".

Here's what I remember from it:

  1. First one defines pairs $\langle a,b\rangle = \{\{a\},\{a,b\}\}$, with this definition one can define $A\times B$ as the set of all pairs $(a,b)$ with $a\in A$ and $b\in B$. However, that's not a good way to proceed, because of the problems you note.
  2. With this definition one defines $\prod_{i\in I} X_i$ as the set of functions $f\colon I \to X_i$ such that $f(i)\in X_i$, here $\{X_i \mid i\in I\}$ is a collection of sets (in other words a function $I\to \mathcal P(\cup X_i)$).
  3. In particular $A^2$ is the set of functions $2\to A$, where $2 = \{\varnothing, \{\varnothing\}\}$, and $A\times B$ is the set of functions $f\colon 2 \to A\cup B$ where $f(\varnothing)\in A$ and $f(\{\varnothing\})\in B$.
  4. Now we forget about that first definition, and proceed with the latter. (Even though we use the former definition to state the latter!) The practical advantage is that now in fact $A\times B\times C$ is actually well-defined, just like any other product, no matter how large the index-set $I$.
  5. It is still not true that $(A\times B)\times C = A\times (B\times C)$, and in fact both are still different from $A\times B\times C$. However, there are bijections between these three sets that are so obvious that for all practical purposes one may consider them to be equal.
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