Surjective map from a countable set to [0, 1]

Question

Consider the map $\varphi$ from $A = \prod_{n = 1}^{\infty}\{ x \in \mathbb{Z} \mid 0 \leq x \leq 9 \}$ to $[0, 1]$ such that $$ \varphi : (a_{1} , a_{2}, \dots) \mapsto 0.a_{1}a_{2}\dots $$ Since $A$ is countable, and $\varphi$ is surjective (every real number in $[0, 1]$ has a decimal representation), $[0, 1]$ is also countable, which is not the case, what's wrong with the reasoning?

Edit: $A$ is uncountable. I thought that countable union of countable sets is countable, however in this case it's the countable product of countable set rather than union, it's the assumption that's wrong, not the conclusion.

One could say that the cardinality of $A$ is $10^{\aleph_0}$ which is uncountable (and equal to $2^{\aleph_0}$). — Jeppe Stig Nielsen, Jun 27 '23 at 14:08
Your final remark is incorrect. That is not a countable union. A countable union of countable sets is countable. But here we have countable product instead. — freakish, Jun 28 '23 at 05:42

freakish · Accepted Answer · 2023-06-27T13:47:17.257

Since $A$ is countable

Unfortunately it is not. And this reasoning here can serve as a proof of that, assuming you already know that $[0,1]$ is uncountable.

If you don't, then the standard Cantor's diagonal argument applies. Assume that there is a sequence $f_{n}$ that covers whole $A$. So it is a sequence of sequences and thus $f_{n,m}$ makes sense.

Define $$a_k=(f_{k,k}+1)\text{ mod }10$$ There is nothing special about this formula. The point is that $a_k\neq f_{k,k}$. I've added mod only to stay in digits set. Any such choice will work. I leave as a simple exercise that $(a_k)$ cannot be in the image of $f_n$.

Surjective map from a countable set to [0, 1]

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