Originally Posted by

**novice** Definition of 0ne-one function:

$\displaystyle f:A\rightarrow B$ is one-one if $\displaystyle f(a)=f(b)$ implies $\displaystyle a=b$.

Now let $\displaystyle A=\varnothing$ and $\displaystyle B=\varnothing$.

I know the logical statement $\displaystyle g: a\notin \varnothing \Rightarrow b\notin \varnothing$ is true, since were $\displaystyle P$ is true and $\displaystyle Q$ is true, $\displaystyle P \Rightarrow Q$ is true by vacuous proof, but it's not the definition of one-one function.

Question: How can we prove $\displaystyle g:\varnothing \rightarrow \varnothing $ using the definition?

$\displaystyle g$(blank)=$\displaystyle g$(blank) implies blank = blank or

$\displaystyle g(\qquad)=g(\qquad)$ implies $\displaystyle \qquad\qquad\qquad$ =$\displaystyle \qquad\qquad\qquad$ ?