The set theoretical reason for $f(\emptyset) = \emptyset$

Question

Why is it that $f(\emptyset) = \emptyset $? I could not write a rigorous explanation for this.
Could anyone explain this for me please?

Could you please tell us what is $f$ and what is $\phi$? – Kenny Wong May 21 '17 at 20:10 — Kenny Wong, May 21 '17 at 20:10
Are you writing the empty set? – Michael Burr May 21 '17 at 20:11 — Michael Burr, May 21 '17 at 20:11

score 11 · Accepted Answer · answered May 21 '17 at 20:14

If you're trying to write $f(\emptyset)=\emptyset$ (i.e., the empty set and not the Greek letter $\phi$), then:

In general, $f(C)=\{f(a):a\in C\}$, in other words, the set of all outputs of $f$ on elements of $C$.

In the case of the emptyset, $f(\emptyset)=\{f(a):a\in\emptyset\}$. Since $a\in\emptyset$ is always false, there are no $a$'s that satisfy the condition, so there are no $f(a)$'s in the set. Hence, the set is empty.

Alternately, you can see this via contradiction, if $b\in f(\emptyset)$, then $b=f(a)$ for some $a\in\emptyset$, but since there are no $a$'s such that $a\in\emptyset$, there is no $a$ for which $b=f(a)$.

The set theoretical reason for $f(\emptyset) = \emptyset$

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