Does 'uncountable sequence' make sense?

In the question Can the natural number have an uncountable set of subsets? the poster uses the phrase 'uncountable sequence'. Does that phrase make sense? No one in the question seems to object, but to me a sequence (which is just a function $f : A \subseteq \mathbb{N} \mapsto X$) seems inherently countable. What am I missing?

• In set theory, a sequence is usually a function whose domain is a well-ordered set for instance an ordinal. Sometimes, a function whose domain is a linear ordering may also be called a sequence. – William Mar 24 '14 at 9:55
• (+1) I wanted to ask that same question as a comment in the original thread - but being completely uneducated in "set-theory" and "set theoretic terminology"i decided to stay away. Nice to see that this is not only a complete individualistic sense/feeling... – Gottfried Helms Mar 24 '14 at 10:05
• @Gottfried: I actually had the opposite problem as an undergrad. I never understood why we insist on limiting ourselves to things indexed by $\Bbb N$ all the time. :-) [I guess that's why I grew up to be a set theory student...] – Asaf Karagila Mar 24 '14 at 10:33
• @Asaf - nice. And your answer gives some really new idea to me; well, naturally: if we think about "gneralizations" as "natural" (which I consider it is). Something to chew on today - thank you! – Gottfried Helms Mar 24 '14 at 11:12
• The idea of "transfinite sequence" (of ordinals or cardinals) is common in set theory. In real analysis one can have transfinite sequences of functions, in topology one can have transfinite sequences (special case of nets/filters) as this paper or this paper illustrate, and here one can even find transfinite sequences of subgroups. – Dave L. Renfro Mar 24 '14 at 18:19

Sequences in "usual mathematics" usually mean to be indexed by $\Bbb N$.

But the only defining property of a sequence is that it is a function from a well-ordered domain, so the notion of "the next term" makes sense. There are uncountable well-ordered sets, and so we can talk about longer sequences, uncountably long sequences. In fact, if you really think about it, a sequence is just a function from an index so, so we don't even care about the ordering either.

The point, however, which seems to confuse you is the fact that sequences are not only in the context of a sequence of real numbers or a sequence of functions, or so on. So we don't care about summability or integrability or limits or whatnot. It's just a convenient way talking about a collection of sets.

In the real numbers we reduce to talking about countable sequences because if we sum (like as series) uncountably many positive real numbers, the result is infinite. Always. So there's no reason to talk about more, and as a result we usually don't care about uncountably sequences of functions either, because we want to consider their sums and limits and so on (even pointwise), and that would be meaningless otherwise.

Well, you can extend that definition and allow any ordinal number in place of $\mathbb{N}$. Or you can drop the well-ordering requirement, and define a $I$-sequence in $X$ as a pair $(f \,:\, I \to X,<)$ of a function $f$ from $I$ to $X$ and a linear order $<$ on $I$. This is equivalent to defining an $I$-sequence on $X$ as a family of elements of $X$ with index set $I$ - usually written $(x_i)_{i\in I},x_i \in X$ - plus an linear order $<$ on $I$.

How usefull that is depends on what you intent to do with it. If you set $X = \mathbb{R}_{\geq 0}$, then you can define $$\sum_{i \in I} = \sup_{F \subset I, F\text{ finite}} \sum_{i\in F} x_i$$ which coincides with the usual definition of the sum of a sequence for countable $I$. Though you can show that such a sum diverges to $\infty$ unless only countably many $x_i$ are non-zero, which restrict the usefullness of this somewhat.