The image of $\emptyset$ under $f$ is empty.

Question

Let A= $\{ \emptyset,\{1\},\{2,\},...,\{n\} \}$. Define $f\colon A \to \mathbb{N}$ such that $f(\emptyset)=0$ and $f(\{n\})=n$.

Is this a counterexample that the image of the $\emptyset$ under an arbitrary function $f$ is empty?

Answer 1

You are defining a function on a set $X$ that has $\emptyset$ as an element. So $f(\emptyset)$ can be anything you like.

But you're probably thinking of the notion of "image of a subspace under $f$".

I (and many texts/papers) always separate these notions notationwise, using square brackets so $f[\emptyset] = \emptyset$ which is indeed always true, and in general $$f[A]= \{y : \exists x \in A: f(x) = y\}$$ is used for the set of $f$-images of elements of $A$. Clearly, if $A$ has no elements, there are also no $f$-images. Some texts write $f(A)$ for $f[A]$ but this can get confusing, as witnessed by your question.