# Disprove : $\varnothing$ from a non-empty set to a non-empty set is a function

Counter-example : Take $A=\{1\}$ and $B=\{2\}.$ Then under $\varnothing$, the element $1$ of $A$ does not have an image in $B$. Thus $\varnothing : A \to B$ is not a function.

I seek feedback about my approach whether it is right or wrong.

I know that $\varnothing$ is a function when $A$ is empty. Otherwise it is not. I have gone through this and this, it helped me to understand the cases when $A$ is empty and when $B$ is empty but $A$ is non-empty.

EDIT : The definition of function that I am using is as follows,

A relation $F$ from a set $A$ to set $B$ is said to be a function if and only if-

1. For every $x \in A$, $\exists y \in B$ such that $(x,y) \in F$,

2. If $(x,y) \in F$ and $(x,z) \in F$, then $y=z$.

• I see no problem here. – smb3 Aug 1 '17 at 4:42

The title can be read two ways. One is that $\emptyset: A \to B$ is not always a function when $A,B$ are non-empty. Your example proves that. Another is that $\emptyset: A \to B$ is never a function when $A,B$ are non-empty. One example cannot prove that. You would then have to argue along the lines of "Take $a \in A$, which we can do because $A$ is non-empty. There is no ordered pair in $F$ with $a$ as a first element, so it fails criterion $1$."
• Yes yes. I actually was also thinking now that I just disproved that "$\varnothing : A \to B$ is a function" by taking counter-example. Thanks. – Error 404 Aug 1 '17 at 5:09