4
$\begingroup$

It seems to me that tuples are just as elementary as sets. Can tuples be reduced to sets?

$\endgroup$

1 Answer 1

10
$\begingroup$

Yes, they can! They are done inductively. First, ordered pairs are defined as $(a, b) = \{\{a\}, \{a, b\}\}$ (exercise: show that $(a, b) = (c, d)$ iff $a = c$ and $b = d$). Then, an $n + 1$ tuple $(x_1, \dots, x_n, x_{n + 1})$ is defined as $((x_1, \dots, x_n), x_{n + 1})$. Lastly, a (countably) infinite sequence, that is, an "infinite"-tuple in some sense, is defined as a function $f: \mathbb{N} \rightarrow X$, where $X$ is some set which contains every coordinate we want. The intuition for this is that I want $f(n)$ to define my $n$th coordinate.

$\endgroup$
12
  • 3
    $\begingroup$ @TorstenSchoeneberg: A function is a set of ordered pairs, no? $\endgroup$
    – JonathanZ
    Aug 26, 2019 at 3:13
  • 3
    $\begingroup$ Absolutely, and it's worth mentioning. I don't like doing it though, mainly because it means that 2-tuples and ordered pairs are different at the set level. This doesn't matter in the long run, as both definitions fulfill the central property of ordered pairs (they're equal iff the coordinates are equal), but it still bothers me aesthetically. $\endgroup$ Aug 26, 2019 at 3:14
  • 1
    $\begingroup$ Thanks, it's amazing to me how non-ordered sets can define ordered tuples like that. And I guess that means {a, b} x {c, d} = {(a, c), (a, d), (b, c), (b, d)} = {{{a}, {a, c}}, {{a}, {a, d}}, {{b}, {b, c}}, {{b}, {b, d}}}? $\endgroup$
    – csp2018
    Aug 26, 2019 at 11:39
  • 1
    $\begingroup$ @csp2018 No. You can see why it won't be this if you consider $(a, b, a)$ and $(a, a, b)$. Under that definition, $(a, b, a) = \{\{a\}, \{a, b\}, \{a, b, a\}\} = \{\{a\}, \{a, b\}\} = \{\{a\}, \{a, a\}, \{a, a, b\}\} = (a, a, b)$ which is a big deal if $a \neq b$. $\endgroup$ Sep 5, 2019 at 3:02
  • 1
    $\begingroup$ Instead, $(a, b, c) = ((a, b), c) = \{\{(a, b)\}, \{(a, b), c\}\} = \{\{\{\{a\}, \{a, b\}\}\}, \{\{\{a\}, \{a, b\}\}, c\}\}$. $\endgroup$ Sep 5, 2019 at 3:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .