Let us think about the empty function $f:\emptyset\rightarrow X$ ($X$ is an arbitrary set.) .

My idea is $f$ is always injective. Iff $X=\emptyset$, $f$ is surjective (so bijective).


Definition of injective is :$x\neq x'\rightarrow f(x)\neq f(x')$. The empty set has no element, so $x\neq x'\rightarrow f(x)\neq f(x')$ is always true.

Definition of surjective is : $\forall y\in Y$, there exists $x\in X$ such that $f(x)=y$.

Iff Y is the empty set,there is no element of $Y$, so $\forall y\in Y$, there exists $x\in X$ such that $f(x)=y$ is true.

  • 1
    $\begingroup$ If "idea" is to mean more than "guess", why not give us your reasoning? $\endgroup$ Mar 17 '20 at 11:16
  • $\begingroup$ Hint: How many elements of the empty set are mapped to a given element of $X$? $\endgroup$
    – celtschk
    Mar 17 '20 at 11:17
  • $\begingroup$ I tried to reason this. Is this O.K.? $\endgroup$
    Mar 17 '20 at 11:28
  • $\begingroup$ Yov've got the definition of injective wrong! $\endgroup$ Mar 17 '20 at 11:44
  • $\begingroup$ @PeterSmith Thanks, I edited. Is tihs O.K.? $\endgroup$
    Mar 17 '20 at 11:48

You are correct. Since $\forall x, y \in \emptyset f(x) = f(y) \Longrightarrow x = y$ is vacuously true, $f$ is always injective.

Now, let's check surjectivity (namely $\forall x \in X\ \exists y \in \emptyset, f(y) = x$). $f$ is:

  • not surjective if $X$ contains any element $x$ since $\exists y \in \emptyset, f(y) = x$ is false.
  • surjective if $X = \emptyset$. Again, the sentence $\forall x \in \emptyset \ \exists y \in \emptyset, f(y) = x$ is vacuously true.

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