5
$\begingroup$

I recently came across the Kuratowski definition of ordered pairs (https://en.wikipedia.org/wiki/Ordered_pair#Defining_the_ordered_pair_using_set_theory).

I'm a bit confused about how the coordinates can be extracted. According to Wikipedia, if we have an ordered pair $p = (a,b) = \{\{a\}, \{a,b\}\}$, we can extract the first coordinate with $a = \bigcup \bigcap{p}$ where $\bigcup$ and $\bigcap$ are arbitrary unions and intersections respectively.

According to the definition of an arbitrary intersection, $\bigcap(\{\{a\}, \{a,b\}\}) = \{a\}$. $\bigcup\{a\}$ however seems a bit problematic. For the definition of an arbitrary union to work $a$ would have to be a set. What if $a$ is e.g. a natural number? A similar issue occurs with extracting the second pair.

Perhaps I'm being overly pedantic or does the definition just automatically assume that $a$ and $b$ are sets?

$\endgroup$
  • 2
    $\begingroup$ This definition of ordered pair is used in the context of a set theory like $\mathsf{ZF(C)}$, in which everything is a set. $\endgroup$ – Brian M. Scott Oct 23 '15 at 22:18
6
$\begingroup$

This is not problematic when you remember that in modern set theory, everything is a set (even the natural numbers!). In particular $a$ is a set.

So $\bigcup\{a\}$ makes sense again, and it is not hard to see why you get $a$.

If you want to take some things as non-sets, then you need to define some class-functions $\varphi_1(z,x)$ and $\varphi_2(z,y)$ which essentially say that $z$ looks like a Kuratowski ordered pair, and $x$ is the unique element which appears in $\bigcap z$; and $\varphi_2$ would say a similar thing about how you extract $y$ from $z$.

$\endgroup$
  • $\begingroup$ Oh I see. Thanks for the clarification! $\endgroup$ – user203664 Oct 23 '15 at 22:21
2
$\begingroup$

In this context everything, including natural numbers, functions, Hilbert spaces, etc., gets encoded as a set, just as in this case an ordered pair gets encoded as a set.

It then becomes tempting to say that everything is a set. That statement is ok if you restrict everything to its proper context. This way of encoding things serves some purposes but it's not necessarily sacred or the last word for all eternity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy