# Why is $A^0 = \{ \emptyset \}$?

Let $A$ be a set and denote $\underbrace{A \times A \times \cdots \times A}_\text{n times}$ by $A^n$. Why is it that $A^0 = \{ \emptyset \}$ ?

By your definition, $A^0$ should be the set of all tuples of length zero of elements of $A$. There is exactly one zero-tuple. One might want to call it $()$, apparently the source that you use calls it $\emptyset$, which is justified for the reasons that Asaf has given.

Because $A^B$ is the set of all functions from $B$ into $A$. Given that $0$ is $\varnothing$, how many functions are there from the empty set into $A$?

• It might pay to expand this a drop to indicate that the usual interpretation of a natural number in this context is a von Neumann ordinal, and perhaps to note that the $\varnothing$ in the result is a function represented by its graph. – dfeuer Dec 12 '13 at 4:32

Asaf Karagila has given the technical answer, so I will give a more hand-wavy answer.

In the real numbers, for example, we want $x^{m+n} = x^{m}\cdot x^{n}$ to hold, which means that we are forced into saying that $x^{0} = 1$. For Cartesian products, we also want $A^{m+n} = A^{m} \times A^{n}$ to hold, and so we are forced to say that $A^{0} = \left\{\varnothing\right\}$.

In order to see that this convention makes sense, just notice that, as sets, $A$ and $A\times\left\{\varnothing\right\}$ are essentially the same, since there is an obvious bijection sending $a\mapsto\left(a,\varnothing\right)$. So we can say, loosely, that they are equal. Technically, they are different, but the same is true that, technically, $A^{m+n}$ is different from $A^{m}\times A^{n}$.

We can't have, say, $A^{0} = \varnothing$ since $A\times\varnothing$ is always $\varnothing$, but that's also why we can't say that $x^{0} = 0$ in $\mathbb{R}$.