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Definition: A set $A$ is countable if it is finite or if there is a bijection $c: \mathbb N \to A$; otherwise it is uncountable.

Let $I_n = \{i\in\mathbb N\mid i \leq n\}$. Then $\prod\limits_{n \in \mathbb N}I_n$ is uncountable.

This question arose when I tried to prove the above theorem.

I found that there exists an injection $\begin {array}{l|rcl} f : & \mathbb N & \longrightarrow & \prod\limits_{n \in \mathbb N}I_n \\ & n & \longmapsto & f(n) \end{array}$

$f(n)$ is defined by $f(n)(i)=\begin{cases}i & \text{ if } i<n\\ n & \text{ if } i \ge n \end {cases}$

I'm not sure if there is a surjection from $\mathbb N$ onto $\prod\limits_{n \in \mathbb N}I_n$?

Akira
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    Why would there be a surjection? – Asaf Karagila Aug 30 '18 at 13:12
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    There is a natural injection from ${1,2}^\mathbb{N}$ into your set, and the former set is uncountable. So the answer is No. – Hans Engler Aug 30 '18 at 13:12
  • A subset of your set is ${(0,x_1,x_2,\ldots) : x_j\in{0,1}\forall j}$. From this set you get a surjection to the set of binary representations of the numbers in $[0,1)$. – amsmath Aug 30 '18 at 13:18

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