For $ n\to\infty $, does $A_{n\to\infty}$ contain all possible infinite length sequences?

Question

I don't claim to have chosen the right words and notations in the question. I'm trying to understand the concepts.

Let, the set $A_n$ is given.

The set of $A_n$, is the set of all possible sequences, which is consist of elements $\left\{0,1,2\right\}$, digit's lentgh equal to $n.$

Example:

$A_3:=\left\{(0,0,0);(0,0,1);(0,0,2);(0,1,0);(0,1,1);\\(0,1,2);(0,2,0);(0,2,1);(0,2,2),(1,0,0);(1,0,1);\\(1,0,2);(1,1,0);(1,1,1);(1,1,2);(1,2,0);(1,2,1);\\(1,2,2);(2,0,0);(2,0,1);(2,0,2);(2,1,0);(2,1,1);\\(2,1,2);(2,2,0);(2,2,1);(2,2,2)\right\}$

Now, if we include $n\to\infty$, (infinite lentgh of sequences) can we say:

Question-1

For $n\to\infty$, the set of $A_{n\to\infty}$ contains all possible infinite length sequences. Is this claim correct?

Question-2

For $n\to\infty$, since the set of $A_{n\to\infty}$ contains all possible infinite length sequences, this imply the cardinality of set of $A_{n\to\infty}$ is equal to $2 ^ {\aleph_0}.$ Is this claim correct?

Thank you very much.

Answer 1

As discussed elsewhere on this page, there's some inherent ambiguity about the use of $\infty$. It is sometimes clear from context, but it is sometimes not.

You could say that $\infty$ here represents $\Bbb N$, or $\omega$ as it is usually denoted in set theory, as the smallest infinite ordinal. And that would indeed make sense. And indeed, in that case, $A_\omega$ would be all the functions from $\Bbb N$ to $\{0,1,2\}$, and yes that would be a set of size $2^{\aleph_0}$.

But you could also think about somehow the limit of the finite sequences, and these would be the sequences which are finite, or if you prefer to think about infinite sequences, then those would be the sequences which are eventually zero, or eventually constant. In this case, the collection of sequences is countable. We can see that by first enumerating the prime numbers, $p_0,p_1,\dots$ and then mapping the [eventually $0$] sequence $a_n$ to $\prod p_i^{a_i}$, as only finitely many $a_i$'s are non-zero, this would be a well-defined natural number and it is easy to verify this is an injection (not a bijection, since $a_i$ is at most $2$).

Answer 2

When you write $A_{n\to\infty}$ I believe that you mean $\lim_{n\to\infty}\bigcup_{k=1}^n A_k$, that is the union of all possible $A_k$.

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The claim in question 1 is incorrect, $\lim_{n\to\infty}\bigcup_{k=1}^n A_k$ only contains finite sequences. It contains an infinite amount of them, with ever growing sizes, but each individual sequence is finite.

Similarly the claim in question 2 is incorrect, based on the false assumption in question 2. The cardinality of $\lim_{n\to\infty}\bigcup_{k=1}^n A_k$ is still $\aleph_0$. It's a bit trickier to come up with a direct mapping, but here's a hint: a countable combination of countable objects is still countable because bijections between $\mathbb{N}^2$ and $\mathbb{N}$ exist.