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For countable sets, what is the advantage of using an indexing set, such as $i \in I$, compared to just using the naturals and the normal enumeration of $1, 2, 3, \ldots$? To me it seems they equivalent unless the sets are uncountable, in which you cannot use something like $\mathbb{N}$ to index your set.

Andrés E. Caicedo
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CBBAM
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    I don't think there's a significant advantage. It might seem aesthetically displeasing to order the indices arbitrarily, if there's not a natural way to order them. – littleO Sep 16 '20 at 19:00
  • You can arrange things more neatly by doing so when the contents of the indexing set are more important than the size of the indexing set. For example, $\det(A)=\sum\limits_{\sigma\in S_n}\left(\text{sgn}(\sigma)\prod\limits_{i=1}^n A_{i,\sigma(i)}\right)$, indexing over the elements in $S_n$ rather than worrying about how to label the elements in $S_n$ from $1$ to $n!$ – JMoravitz Sep 16 '20 at 19:02
  • @littleO So if I were to somehow index a finite number of elements in an uncountable infinite set, how would I do that in terms of notation? For example, if I were to index the $n$ arbitrary irrationals in $\mathbb{R}$. – CBBAM Sep 16 '20 at 19:03
  • @CBBAM You don't try to force a different indexing on it, you just do it the natural way: if the set of length-$n$ ordered tuples of irrationals is denoted $(\mathbb{R}\setminus\mathbb{Q})^n$, and you can just use that as your index set. – Noah Schweber Sep 16 '20 at 19:07
  • There isn't much of a difference - for any enumeration you do with an index set, you can do the same such enumeration with the natural numbers. This is the definition of countability. – rubikscube09 Sep 16 '20 at 19:12

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Even ignoring the advantage of avoiding a notational disagreement between countable and uncountable cases, just because something is countable doesn't mean that there is a canonical bijection between it and $\mathbb{N}$. By being flexible about index sets we can make sure that the Cartesian products we build don't have (or require us to input any!) any "extraneous information."

To see how this might get used in an argument, consider for example the following:

$(*)\quad$ Given a countable set $S\subseteq\mathbb{R}$, let $\mathcal{F}$ be all the ways of assigning to each $s\in S$ some $t<s$.

For each $S$ this is a Cartesian product; specifically, it's $$\prod_{s\in S}\{t\in\mathbb{R}: t<s\}.$$ Now if we really want to, since each $S$ is countable we could instead look at $$\prod_{n\in\mathbb{N}}\{t\in\mathbb{R}: t<f(n)\}$$ for some fixed bijection $f:\mathbb{N}\rightarrow S$, but this adds a layer of complexity: we need to fold in some choice of $f$, and in general there won't be a "best" choice to make here. In particular, if we insist on doing this then the definition of $\mathcal{F}$ given in $(*)$ is incomplete for our purposes since it doesn't tell us which $f$ to pick.

Noah Schweber
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