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Consider an arbitrary set $A := \{ a_1, a_2, a_3, ... \}$ of cardinality $ \aleph_0 $. I'm trying to determine if the set of functions $F:= \{ f \, | \, f : A \rightarrow \{0,1\} \}$ is countably infinite. I don't see an obvious bijection to the natural numbers which leads me to think it might not be. If so, then it seems like you could construct a diagonalization argument based on introducing a $g$ that differs from some listed $f_n$ by only one out-put -- taking $g(a_n)= 1- f_n(a_n)$ to arrive at a contradiction. Is that the general idea? Guidance would be appreciated.

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    Do you mean ".. of cardinality $\aleph_0$"? Writing $\aleph$ without the zero subscript is sometimes used to mean the cardinality of $\mathbb R$. – 2017-02-02
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    Your set $F$ is essentially the powerset of $A$, so by a well known argument such as the one you've provided, it's not countable – 2017-02-02

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The set of functions $F$ is essentially analogous to the set of real values between $0$ and $1$, expressed in binary. (It's the set of reals in the unit interval because $A$ is infinite.) So $F$ is uncountable.

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    Interesting. How would you construct the bijection between them, more formally? – 2017-02-03
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    @cloudchamber Do you know how to produce a bijection between the powerset of $\mathbb{N}$ and $(0, 1)$? If so, it's exactly the same. If not, here's the idea: take an $f$, and look at the real with $f$ as its binary expansion. This won't quite be injective, but it's fixable. – 2017-02-03
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Yup, the set is uncountable, and the proof you've described is correct.