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This exercise is from my course textbook and comes with a bunch of other exercises which practice the theorem that countable union of countable sets is countable.

So I started by notating for every $k\in \mathbb{N}$ the set $X_k$ as the set of all periodic sequences from the index $k$ of natural numbers. now, $X=\displaystyle{\bigcup_{k\in \mathbb{N}}{X_k}}$ so it's only left to prove that for each $k\in \mathbb{N}$ the set $X_k$ is countable but I can't find an injection from $X_k$ to a countable set.

Vegetal605
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  • Hint: http://math.stackexchange.com/questions/200389/show-that-the-set-of-all-finite-subsets-of-mathbbn-is-countable – lulu Sep 01 '15 at 15:23

2 Answers2

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Hint: Set $X_k^n$ to be the set of integer sequences that have period $k$, starting at index $n$. Now observe that $X = \bigcup \limits_{(k, n) \in \mathbb{N}^2} X_k^n$.

Dominik
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  • Do you mean the bijection which sends the first $k$ elements of a sequence in $X_k$? – Vegetal605 Sep 01 '15 at 15:00
  • Yes. This solutions works assuming you know that the cartesian product of two (or more generally: finitely many) countable sets is countable. – Dominik Sep 01 '15 at 15:03
  • This was my first guess but I don't see how that's work here. since I don't know what the period of the sequence I can't restore the original sequence in $X_k$ given it's first $k$ elements. – Vegetal605 Sep 01 '15 at 15:05
  • I assumed that with "index" you meant the period of the sequence. Then the sequence is completely determined by the first $k$ values. Note that e.g. $a_1 = a_{1 + k}$, $a_2 = a_{k + 2}$ etc. – Dominik Sep 01 '15 at 15:07
  • actually since I can't find the injection I need, I thought that maybe what you just said is the meaning of the question. the formulation of the question was exactly how I wrote it in the title. so how should i interpret it? – Vegetal605 Sep 01 '15 at 15:10
  • for example, does the sequence $(1,2,7,3,5,9,6,5,9,6,5,9,6,\ldots )\in X$? – Vegetal605 Sep 01 '15 at 15:12
  • Now I understand the question, the sequence is meant to be periodic, starting from a certain index. This menas there is an $N \in \mathbb{N}$ so that $a_{n} = a_{n + k}$ for all $n \ge N$, where $k$ is the period of the sequence. I've changed my hint. – Dominik Sep 01 '15 at 15:16
  • Thanks! now it makes sense to me. – Vegetal605 Sep 01 '15 at 15:17
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For $k \ge 1$, call $U_k = \{ \mbox{sequences } (x_n)_n \in X : \max_n x_n =k-1 \}$. Clearly $$X= \bigcup_k U_k$$ so it is enough to show that $U_k$ is countable. Clearly $U_1$ has only one element : the constant $0$ sequence.

For $k \ge 2$, do as follows: inject $U_k \longrightarrow \Bbb{Q}$ with the map $$(x_n)_n \mapsto \sum_{n=0}^{\infty} x_n (k+1)^{-n}$$ i.e. you associate to a periodic sequence its periodic $(k+1)$-nary number, which is rational.

Maybe the most natural choice would have been $(x_n)_n \mapsto \sum_{n=0}^{\infty} x_n k^{-n}$. But the problem is that $k$-nary representation of rational numbers is not unique (problems arise for example when you have $0.49999999\dots = 0.50000000\dots$). Avoiding the largest cipher, we get uniqueness of representation.

Finally, since $\Bbb{Q}$ is well-known to be countable, you can conlude that $U_k$ is countable.

Crostul
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