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When showing that two sets $A$ and $B$ have the same (finite or infinite) cardinality, it is usually done by constructing a function and showing that it is a bijection. However, in some cases, constructing such a bijection is not always easy. This can be worked around by constructing two functions, in a number of different combinations, for instance: $\phi:A\rightarrow B$ and $\psi:B \rightarrow A$ both injective or both surjective.

My question is, in the cases where it's easier, why is this not done? An example would be between the Cantor set and the closed unit interval. Constructing a bijection means taking great care, since any number with trailing zeros can also be written with trailing $($base $-\:1)$s.

Constructing an injection each way is easy, though. The Cantor set can by construction be seen as a subset of the unit interval, so there's inclusion. Any number in the unit interval has a binary expansion, so pick one of the expansions for each number and this will give an element in the cantor set by going left and right for 0 and 1.

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    I often see this done, i.e. every proof I've seen that the Cantor set has cardinality $\mathfrak c$.2012-08-12

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One note: depending on whether you're taking the Axiom of Choice for granted, $\phi,\psi$ both being surjective isn't enough to guarantee a bijection. Schroeder-Bernstein Theorem tells us that it is enough if $\phi,\psi$ are both injective, though.

I more often find myself constructing two injections than one bijection, personally, and many texts do the same.

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If all you seek is that the cardinality is the same, then that is fine. However, the bijection you get by apply the proof the Schröder-Bernstein Theorem may not be very nice.

Sometime the explicit bijection is necessary. For example, if you want to know if $2^\mathbb{N}$ is homeomorphic to the standard Cantor Set, you will need an explicit bijection and show continuity of this function and its inverse.

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    +1: That's a *great* example of an instance where a bijection is necessary!2012-08-13