Show that the set $${\mathbb N}^{<{\mathbb N}} = \cup _{k=0}^{\infty } {\mathbb N}^{\{0, 1, \ldots , k-1\}}$$ of all finite sequences of natural numbers is countable.
Here are two standard ways to see that the set of all finite sequences of naturals
$$ \mathbb{N}^{<\mathbb{N}} \;=\;\bigcup_{k=0}^\infty \mathbb{N}^{\{0,1,\dots,k-1\}} $$
is countable.
1. Union of countably many countable sets
-
Each fixed‑length piece is countable. For each $k\ge1$, the set of $k$‑tuples
$$ \mathbb{N}^k = \{\, (n_0,n_1,\dots,n_{k-1}) : n_i\in\mathbb{N}\} $$
is in bijection with $\mathbb{N}$. For example, one can use the usual Cantor–Gödel pairing repeatedly to pair off coordinates:
$$ \pi:\mathbb{N}\times\mathbb{N}\;\xrightarrow{\sim}\;\mathbb{N}, \quad (a,b)\;\mapsto\;\tfrac12(a+b)(a+b+1)+b, $$
and then pair $\pi(\pi(n_0,n_1),n_2)$, etc., to get a bijection $\mathbb{N}^k\cong\mathbb{N}$.
-
A countable union of countable sets is countable. We have
$$ \mathbb{N}^{<\mathbb{N}} \;=\; \bigcup_{k=0}^\infty \mathbb{N}^{k} \,, $$
and since each $\mathbb{N}^k$ is countable, their (countable) union is again countable.
Hence $\mathbb{N}^{<\mathbb{N}}$ is countable.
2. Explicit injective coding via prime factorisation
Define a map
$$ \Phi:\mathbb{N}^{<\mathbb{N}}\;\longrightarrow\;\mathbb{N}_{>0} $$
by sending a finite sequence
$$ s=(n_0,n_1,\dots,n_{k-1}) $$
to the integer
$$ \Phi(s) \;=\; 2^{n_0}\,\cdot\,3^{n_1}\,\cdot\,5^{n_2}\,\cdots\,p_k^{\,n_{k-1}} \quad \bigl(p_k\text{ = the }k\text{th prime}\bigr). $$
- Injectivity. By the Fundamental Theorem of Arithmetic, every positive integer has a unique prime‑power factorisation; in particular, different sequences $s$ give different factorizations, so $\Phi$ is injective.
- Conclusion. An injective map from $\mathbb{N}^{<\mathbb{N}}$ into $\mathbb{N}$ shows $\#\mathbb{N}^{<\mathbb{N}}\le\#\mathbb{N}$. Since $\mathbb{N}^{<\mathbb{N}}$ is infinite, its cardinality is exactly $\aleph_0$.
Either argument establishes that the set of all finite sequences of natural numbers is countable.
Note: ChatGPT’s uses of \N
, which rendered correctly in its own UI, were changed into \mathbb{N}
here.
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