How does the set of all functions $\{f \,|\, f: \emptyset \to \{0,1\}\}$ look like? Is it empty or does it contain infinitely many functions? Does the definition $f: \emptyset \to \{0,1\}$ make sense at all?

I was wondering because we know that the two sets $\{0,1\}^X$ and $\mathcal{P}(X)$ have the same cardinality. But this is only true if $X$ is non-empty, right?

  • $\begingroup$ What have you tried so far? Have you tried considering the definition of "function"? $\endgroup$ – skyking Dec 11 '17 at 10:03

Yes, it makes sense. There is one and only one function from $\emptyset$ into $\{0,1\}$, which is the empty function. Think about the definition of function as a set of ordered pairs to see why.

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    $\begingroup$ @philmcole If $X=\emptyset$ then $\mathcal P(X)$ has only one element. $\emptyset \in\mathcal P(X)$ but $\{\emptyset\}\notin\mathcal P(\emptyset)$ because $\{\emptyset\}\not\subseteq\emptyset.$ $\endgroup$ – bof Dec 11 '17 at 10:20
  • $\begingroup$ @bof Oh, you're totally right. So the theorem holds also if $X=\emptyset$ since we have $\{0,1\}^X = \{f \,|\, f: \emptyset \to \{0,1\}\} = \{\text{the empty function}\}\sim \{\emptyset\} = \mathcal{P}(X)$ $\endgroup$ – philmcole Dec 11 '17 at 10:36

It is known as the empty function.

For any set $A$, there is exactly one function from the empty set to $A$, namely the empty function:

$$f_A: \emptyset \to A.$$

The graph of an empty function is a subset of the Cartesian product $\emptyset × A$. Since the product is empty the only such subset is the empty set $\emptyset$.

  • $\begingroup$ I have a weird (maybe naif) argument: since '$x \in \emptyset$ implies that any proposition about $x$ is true', can we also say that $x \in f^{-1}(1)$? If we do this for all $x \in \emptyset$ can we conclude that $f$ is constant? And analogously, if we pick $y \in \emptyset$ such that $y \in f^{-1}(0)$ whereas for any other $x \in \emptyset, x \in f^{-1}(1)$, can we conclude $f$ is not constant? This seems to be a contradiction, does not it? $\endgroup$ – Gibbs Dec 11 '17 at 10:12
  • $\begingroup$ related $\endgroup$ – Siong Thye Goh Dec 11 '17 at 10:19
  • $\begingroup$ I read that $f$ is a constant function, but I see no mention of the contradiction in my previous comment... $\endgroup$ – Gibbs Dec 11 '17 at 10:25

A function $f:A\to B$ is a set of ordered pairs $(a,b)$ with $a\in A,b\in B$. In this case there is no $a\in A$ so therefore $f$ is an 'empty function'


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