For any sets $A$ and $B$, the set of all functions from $A$ to $B$ is denoted $\text{}^A B$. If there is a one-to-one function $f : A \rightarrow B$, $A \precsim B$. Suppose $A \precsim B$ and $C \precsim D$. If $A = \varnothing$, can $\text{}^A C \precsim \text{}^B D$ be true?
I think that since A is empty, $\text{}^A C$ should be empty, and $\text{}^A C \precsim \text{}^B D$ would be vacuously true.
Am I wrong?
You're almost right - if $A=\emptyset$, there is exactly one function from $A$ to $C$, namely the empty function (as a set of ordered pairs, it is just $\emptyset$).
This allows the inequality $^AC\le{}^BD$ to fail in case $D$ is empty and $B$ is nonempty - since then $^BD$ is empty!
But, if we assume merely that $D\not=\emptyset$, then $A=\emptyset$ does indeed imply $^AC\le{}^BD$.
In particular, if $A=D=C=\emptyset$ and $B\not=\emptyset$, then $^AC>{}^BD$.
