Which elements does $(A^B)^C$ include?

Question

Suppose we have sets $A,B,C = {(0,1)}$, i.e each set has only 2 elements. Then $(A^B)^C$ should have elements
$0 \to (0 \to 0)$
$0 \to (0 \to 1)$
$0 \to (1 \to 0)$
$0 \to (1 \to 1)$
$1 \to (0 \to 0)$
$1 \to (0 \to 1)$
$1 \to (1 \to 0)$
$1 \to (1 \to 1)$

Am I correct ?

I figured out that $A^B = (f_1, ... , f_4)$. It's elements are
$((0,0),(1,0)),$
$((0,0),(1,1)),$
$((0,1),(1,0)),$
$((0,1),(1,1)),$

@AsafKaragila Could you please then answer, which elements will $(A^B)^C$ include. — Rustem Sadykov, Sep 17 '20 at 06:40
As per answer to your previous post, $A^B$ is the set of functions from $B$ to $A$. This means that an element $f \in A^B$ will be a set of pairs: $f = { (b,a) \mid b \in B, a \in A }$. — Mauro ALLEGRANZA, Sep 17 '20 at 06:53
Following your example, $f= { (0,0), (1,0) }$. It must be a function: thus, we cannot have two different outputs for the same input. — Mauro ALLEGRANZA, Sep 17 '20 at 06:56
Now, what is a function $F : C \to A^B$ ? It will be a function that maps elements of $C$ to elements of $A^B$, i.e. to functions like that previously defined. E.g. $F(0)=f$, etc. i.e. $F = { (c,f) \mid c \in C, f \in A^B }$. — Mauro ALLEGRANZA, Sep 17 '20 at 06:58

freakish · Accepted Answer · 2020-09-17T07:03:28.617

I assume that by $A^B$ you mean the set of all functions $B\to A$. So elements of $(A^B)^C$ are all functions $C\to A^B$. Since each set has only two elements then we can list all those functions. First of all if $f:C\to A^B$ then for any $x\in C$ we have that $f(x)$ is a function $B\to A$. And so we can evaluate $f(x)$ on $y\in B$. And in fact $f$ is uniquely determined by values $f(x)(y)$. And so all possible functions $C\to A^B$ are the following:

$$f_1(0)(0)=0,\ \ f_1(0)(1)=0,\ \ f_1(1)(0)=0,\ \ f_1(1)(1)=0$$ $$f_2(0)(0)=0,\ \ f_2(0)(1)=0,\ \ f_2(1)(0)=0,\ \ f_2(1)(1)=1$$ $$f_3(0)(0)=0,\ \ f_3(0)(1)=0,\ \ f_3(1)(0)=1,\ \ f_3(1)(1)=0$$ $$f_4(0)(0)=0,\ \ f_4(0)(1)=0,\ \ f_4(1)(0)=1,\ \ f_4(1)(1)=1$$ $$f_5(0)(0)=0,\ \ f_5(0)(1)=1,\ \ f_5(1)(0)=0,\ \ f_5(1)(1)=0$$ $$f_6(0)(0)=0,\ \ f_6(0)(1)=1,\ \ f_6(1)(0)=0,\ \ f_6(1)(1)=1$$ $$f_7(0)(0)=0,\ \ f_7(0)(1)=1,\ \ f_7(1)(0)=1,\ \ f_7(1)(1)=0$$ $$f_8(0)(0)=0,\ \ f_8(0)(1)=1,\ \ f_8(1)(0)=1,\ \ f_8(1)(1)=1$$ $$f_9(0)(0)=1,\ \ f_9(0)(1)=0,\ \ f_9(1)(0)=0,\ \ f_9(1)(1)=0$$ $$f_{10}(0)(0)=1,\ \ f_{10}(0)(1)=0,\ \ f_{10}(1)(0)=0,\ \ f_{10}(1)(1)=1$$ $$f_{11}(0)(0)=1,\ \ f_{11}(0)(1)=0,\ \ f_{11}(1)(0)=1,\ \ f_{11}(1)(1)=0$$ $$f_{12}(0)(0)=1,\ \ f_{12}(0)(1)=0,\ \ f_{12}(1)(0)=1,\ \ f_{12}(1)(1)=1$$ $$f_{13}(0)(0)=1,\ \ f_{13}(0)(1)=1,\ \ f_{13}(1)(0)=0,\ \ f_{13}(1)(1)=0$$ $$f_{14}(0)(0)=1,\ \ f_{14}(0)(1)=1,\ \ f_{14}(1)(0)=0,\ \ f_{14}(1)(1)=1$$ $$f_{15}(0)(0)=1,\ \ f_{15}(0)(1)=1,\ \ f_{15}(1)(0)=1,\ \ f_{15}(1)(1)=0$$ $$f_{16}(0)(0)=1,\ \ f_{16}(0)(1)=1,\ \ f_{16}(1)(0)=1,\ \ f_{16}(1)(1)=1$$

resulting in a total of $16$ functions. This agrees with our understanding of cardinalities: $|(A^B)^C|=|A^B|^{|C|}=(|A|^{|B|})^{|C|}$ which evaulates to $(2^2)^2=16$.

There's a pattern here. In fact each function $f:C\to A^B$ has a corresponding $g_f:C\times B\to A$ function:

$$g_f(x,y):=f(x)(y)$$

And this correspondence $(A^B)^C\to A^{C\times B}$, $f\mapsto g_f$ is a bijection, similarly to number exponentiation. And finding all functions $C\times B\to A$ is arguably easier.

Ben Grossmann · Answer 2 · 2020-09-17T07:05:59.017

Recall that $A^B$ contains the set of all functions from $B$ to $A$. In order to define a function $f:B \to A$, we must assign an output $f(b) \in A$ for every element $b \in B$.

With that in mind, let's describe one element from $(A^B)^C$: first, we define the functions $f_0,f_1:B \to A$ as follows: $$ f_0(b) = \begin{cases} 0 & b = 0\\ 0 & b=1,\end{cases} \quad f_1(b) = \begin{cases}1 &b = 0\\0 & b = 1. \end{cases} $$ We then define $F:C \to A^B$ by $F(0) = f_0$ and $F(1) = f_1$.

If you prefer, we could instead specify this function as a collection of ordered pairs: if we consider each element of $A^B$ to be a subset of $B \times A$, then $F$ is the following subset of $C \times \mathcal P(B \times A)$: $$ F = \{(0,\{(0,0),(1,0)\}),(1,\{(0,1),(1,0)\})\}. $$

Note that by "uncurrying", we can think of $F$ as a function $\Phi: C \times B \to A$. In particular, we define $\Phi:\{0,1\} \times \{0,1\} \to \{0,1\}$ by $$ \Phi(c,b) = F(c)(b) = f_c(b). $$ So, for example, $\Phi(1,0) = f_1(0) = 1$. This is often useful because it is easier to think about (or manipulate) a function that has two inputs than it is to think about a function whose output is itself another function.

Written as a subset of $A\times(C \times B)$, $\Phi$ is the set $$ \Phi = \{((0,0),0),((0,1),0),((1,0),1),((1,1),0)\}. $$

Which elements does $(A^B)^C$ include?

2 Answers2