I don't know much but I think there does not exist any injection from higher dimension to lover. Or any surjection from lower to higher. So first 3 options are out. 4th one I'm not sure about.
One one onto functions
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functions
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1Turns out all 4 are true. See for instance https://proofwiki.org/wiki/Cartesian_Product_of_Countable_Sets_is_Countable – 2017-01-28
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0All the sets mentioned in the question are countable, so there is a bijection between any two of them. – 2017-01-28
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0I totally mistook it for something else....haha – 2017-01-28
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0It's reasonable to think there can't be an injection from higher dimension to a lower, but as it turns out that simply isn't the case. A counter example is 1 to (1,1), 2 to (1,2) and 3 to (2,1) and 4,5,and six to (1,3), (2,2)(3,1) and 7 to 10 is (1,4)(2,3)... (4,1) etc. Once you get that, iteration implies all of those are true. – 2017-01-28
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Ok. Here goes a little generalization (maybe) and a hint:
Let $A$ be a countable set then $A'_n = \underbrace{A \times A \times \ldots \times A}_{\text{n times}}$ is also countable. Since countable sets are in bijection with $\mathbb{N}$ so, there exists a bijection from $A$ to $A'_n$ $\forall n \in \mathbb{N}$.
Hint: Try using induction on $n$. (Observe that $A'_1 = A$ so $A'_1$ is countable).