Suppose $X$ is infinite. I think we can show that $|X\times\mathbb{N}| = |X|$. How do we construct an injective function $f:X\times\mathbb{N} \to X$?
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The axiom of choice is necessary here, so you cannot construct an injection like that. However using choice, or one of its many equivalents, we can do it quite easily. – Asaf Karagila Nov 03 '16 at 06:44
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When $X$ is infinite you can partition as $X=\sqcup_{i\in I}X_i$, each $X_i$ being in one-to-one correspondence with $\mathbb{N}$. Let us call $\phi_i : X_i\rightarrow \mathbb{N}$ such a bijection. Then, the "trick" is to construct a bijection between $X_i\times \mathbb{N}$ and $X_i$ which can be done through any bijection between $\mathbb{N}\times \mathbb{N}$ and $\mathbb{N}$.
Duchamp Gérard H. E.
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@Alexis: You cannot construct such a partition, since the axiom of choice is necessary here. – Asaf Karagila Nov 03 '16 at 06:43
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@AlexisOlson Well, starting from such an unframed question, I admit I assumed that we were within ZFC (standard default assumption). Now, you need a form of axiom of choice in order to get the partition, afterwards the remaining part is constructible. – Duchamp Gérard H. E. Nov 03 '16 at 13:00