Let $A, B, X, Y$ be sets with $X \preccurlyeq A$ and $Y \preccurlyeq B$. Prove that, apart from exceptional case(s), $X^Y \preccurlyeq A^B$.

Question

I'm trying to solve the following problem:

Let $A, B, X, Y$ be sets with $X \preccurlyeq A$ and $Y \preccurlyeq B$. Prove that, apart from exceptional case(s), $X^Y \preccurlyeq A^B$.

Here, $X \preccurlyeq A$ means there's an injection from $X$ to $A$ etc, and $X^Y$ is the set of functions from $Y$ to $X$ etc.

Presumably the exceptional cases are going to be where some of $A, B, X, Y$ are empty (and I imagine in a proof you'd see exactly where you need certain sets to be non-empty).

I've played around with the injections from $X$ to $A$ and $Y$ to $B$ to try to construct a new one, but I can't seem to find anything that works. Any help you could offer would be very appreciated.

Answer 1

I guess, we can try something as follows. A function is in fact a functional relation, thus $$ f \in X^Y \Rightarrow f \subset X \times Y; \quad g \in A^B \Rightarrow g \subset A\times B; $$ such that $$ \begin{aligned} \forall (x_1; y_1),(x_2; y_2) \in f&: y_1 = y_2 \Rightarrow x_1 = x_2;\\ \forall (a_1; b_1),(a_2; b_2) \in g&: b_1 = b_2 \Rightarrow a_1 = a_2.\\ \end{aligned} $$

• Consider injections $$ i_X : X \rightarrow A;\quad i_Y : Y \rightarrow B; $$ I can construct mapping $$ i : X \times Y \rightarrow A \times B; \quad i: (x,y) \mapsto (i_X(x); i_Y(y)), $$ and check that the mapping is an injection. Consider $(x_1;y_1) \neq (x_2;y_2)$, this means $x_1 \neq x_2$ or $y_1 \neq y_2$. Thus we get $i_X(x_1) \neq i_X(x_2)$ or $i_Y(y_1) \neq i_Y(y_2)$. Either way we can conclude $(i_X(x_1); i_Y(y_1)) \neq (i_X(x_2); i_Y(y_2))$. Worth noting: property of being injective remains if we restrict $X \times Y$ to any subset.

• Now consider mapping $$ I: P(X \times Y) \rightarrow P(A \times B) $$ that acts as follows $$ \forall S \in P(X \times Y): I(S) = \{i(s)| s\in S\}. $$ Now I claim $I$ is an injection. Consider $S_1 \neq S_2$, which means one of them contains at least one element, say $s$, that the other does not. Without loss of generality suppose $s \in S_1$ and $s \notin S_2$. Using injectivity of $i$ we get $$ s \notin S_2 \rightarrow \forall s' \in S_2: s' \neq s \rightarrow \forall s' \in S_2: i(s') \neq i(s) \rightarrow i(s) \notin I(S_2), $$ which means $I(S_1) \neq I(S_2)$, thus $I$ is injective.

• Now we know that $I$ maps relations injectively. The only unsolved problem is whether it does not "destroy" the property "to be a function". If not -- we are done here as we found an injection from functions $X^Y$ to $A^B$. So let's prove this property. Since $i_Y$ is an injection and $(x;y) \in f$, where $f$ is a function $$ \begin{aligned} \forall (i_X(x); i_Y(y)) \in A \times B: i_Y(y_1) = i_Y(y_2) \Rightarrow y_1 = y_2 \Rightarrow\\ \Rightarrow x_1 = x_2 \Rightarrow i_X(x_1) = i_X(x_2). \end{aligned} $$ The latter is true for any subset of $A \times B$, thus $I$ is injection from functions $X^Y$ to functions $A^B$.