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I want to prove this.

Let $f$ is a bijection between $A$ and $B$,

$A$ is a countably infinite set if and only if $B$ is a countably infinte set.

I use definition A is a countably infinite set then there exist a bijection between A and $\mathbb N$,

but i don't know to show that $B$ is a countably infinite set.

Please, give me a hint or prove this.

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    $\begingroup$ Hint: Bijections are invertible. $\endgroup$ – Jaap Scherphuis Jul 13 '18 at 8:21
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    $\begingroup$ Yes, I see hahaha $\endgroup$ – Chung wow Jul 13 '18 at 8:22
  • $\begingroup$ and composable too. $\endgroup$ – Henno Brandsma Jul 13 '18 at 8:40
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Let $g:A \to B$ be a bijection.

Suppose $A$ is countably infinite so there exists a bijection $f_A: A \to \mathbb{N}$. Then $g^{-1} \circ f_A$ is a bijection from $B$ to $\mathbb{N}$. (inverses and compositions of bijections are bijections). So $B$ is countably infinite

If $B$ is countably infinite, there exists a bijection $f_B: B \to \mathbb{N}$, and then $f_B \circ g: A \to \mathbb{N}$ is a bijection so $A$ is countably infinite.

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If you were able to count $A$ (i.e. assign a distinct number to every element), using the bijection you can transfer all the labels of the elements of $A$ to those of $B$, and $B$ is counted as well.

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