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 $y=z$.

  • 1
    $\begingroup$ I see no problem here. $\endgroup$ – 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 $1$."

  • $\begingroup$ 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. $\endgroup$ – Error 404 Aug 1 '17 at 5:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.