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A sequence is called disjunctive if it contains all sequences within it. More specifically, a sequence is disjunctive in base $b$ if it contains all finite sequences (or base $b$ words) within its base $b$ representation.

This is an interesting property, and it makes me wonder. Do there exist sequences that contain all possible countably infinite sequences? If so, what are they called?

I can't think of a reason they shouldn't exist, but maybe there's a good reason that I'm just overlooking. My first concern was that a sequence containing infinitely many infinite subsequences might already be impossible. But a trivial example shows it's possible:

Consider the set of all countable sequences on the natural numbers which are strictly monotonically increasing. Call it $\mathcal{S}$.

Define $s_o = (1, 2, ...)$ - the sequence which simply enumerates the natural numbers. Clearly, every $s \in \mathcal{S}$ is a subsequence of $s_o$.

$$ \forall s \in \mathcal{S} , s \subseteq s_o $$

And $\mathcal{S}$ contains infinitely many sequences. (Though I'm not sure if $\mathcal{S}$ is countable or not.) So unless I've made a mistake, it's possible for a single countable sequence to contain infinitely many countable subsequences.

But that's essentially a toy problem - I'm a long way off from answering my question and I'm not even sure what terms to search for related results.

kdbanman
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    The answer is no, since there are uncountably many such sequences. – Rushabh Mehta Dec 10 '19 at 02:50
  • Ah, and there can be no countable sequence containing uncountably many elements, let alone uncountably many subsequences. Thanks for the insight! – kdbanman Dec 10 '19 at 03:11
  • Though I wonder, what if one relaxed the conventional definition of "sequence" to include uncountably many elements? (Not trying to be contrarian, just really curious.) To use the language of this answerer, "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." – kdbanman Dec 10 '19 at 03:13
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    The answer is yes. Note that there is an obvious bijection between the set of countable sequences and $\mathbb R$, so consider the set of elements of the form $(r,i)$, where $r$ is the sequence label from $\mathbb R$, and $i$ is the element of the sequence in $\mathbb N$. The dictionary ordering on this set is a well-order (assuming some well-order on $\mathbb R$, of course. AC always find its way into well-order based questions). – Rushabh Mehta Dec 10 '19 at 03:20

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