Let $E$ and $F$ be finite sets. If $card(E)=card(F)\not = 0$ then an injective map from $E$ to $F$ must be bijective. I'm asking why we single out the case of the cardinalities are different from zero?
[Math] An injective map between two sets of the same cardinality is bijective.
elementary-set-theory
Best Answer
There is no reason to do this, but perhaps the reason is because there is a unique map $\varnothing \to \varnothing$, and it is naturally bijective, so there is nothing to say. (If you think of a map $f : E \to F$ as being a subset of the cartesian product $E \times F$ defined by the couples $(e,f(e))$ for $e \in E$, then the graph of the function $\varnothing \to \varnothing$ is simply the empty set.
Edit : Another way to look at it is that the empty set is a subset of every set ("because all its elements are in every set" is vacuously true) so the inclusion map from the empty set to any set exists.
Edit 2 : Even using the theorems in the book you can show that the result is true in the case where $|E|=|F|=0$. Paragraph 4.2 of page 156 says any map $\varnothing \to F$ is injective, and Paragraph 4.4 of page 158 says at the 2) example that any map $\varnothing \to \varnothing$ is bijective. Therefore the (only) map $\varnothing \to \varnothing$, according to the book up to page 158, is injective and bijective.
However, one good reason to exclude the case $|E|=|F|=0$ in the statement of their theorems, is that the proofs they give are not valid in this case. If they wanted to include this case in the theorem, they would have to prove the empty case separately, which they didn't. (I am talking about the proof of 4) at page 171.) The proof goes wrong when they say "suppose $f$ is an injection of $[[1,n]]$ to itself which is not surjective ; thus there exists $y \in [[1,n]]$ such that (...)" and in the empty case this is false.
Hope that helps,