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What is the cardinality of all of the functions $f:R\to \{0,1\}$ so that $\forall {a \notin Q}:\ f(a)=0$

My intuition tells me that the cardinality is $\aleph$, so I was trying to find a bijection to a set with a cardinality $\aleph$, but couldn't really find anything.

Is it the right way to go?

Ѕᴀᴀᴅ
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Lilo
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1 Answers1

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Considering how the function is already determined for irrationals, you only need to define it for rationals. Thus, the question is equivalent to asking the cardinality of the following set of functions:

$$S = \{f:\mathbb{Q} \to \{0, 1\}\}.$$

It is quite standard that the cardinality of $S$ is the same as that of the power set of $\mathbb{Q}$.
(Given a subset $A \subset \mathbb{Q}$, consider the function $\chi_A:\mathbb{Q} \to \{0, 1\}$ such that $\chi_A(q) = 1$ iff $q \in A$. The pairing $A \mapsto \chi_A$ is bijective which can be easily shown.)

It is also standard that the power set of $\mathbb{Q}$ has the same cardinality as that of $\mathbb{R}$. This means that your original intuition was wrong.

Moreover, it is not necessary to explicitly construct a bijection. You could appeal to Schröder–Bernstein theorem which tells you that it suffices to construct injections in both directions.

Also, considering how $|\mathbb{Q}| = |\mathbb{N}|$, it would be easier to work with $\mathbb{N}$ instead. You can find some answers here.

Aryaman Maithani
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  • The function F is total, can I still define S to be only for rationals? – Lilo May 14 '20 at 13:03
  • S is not the set that you were originally looking at. However, you can see that there's a clear bijection between your original set of functions and S. Given any function $f \in S$, simply extend it to a function on $\mathbb{R}$ by defining $f$ to be $0$ on irrationals. – Aryaman Maithani May 14 '20 at 13:06