0
$\begingroup$

It is said that the $\{\}$ is the initial object of the $\mathcal{SET}$ - category of sets and functions.

By definition, it implies that for all $S \in Obj(\mathcal{SET})$ there is exactly one function $f: \{\} \mapsto S$. And that seems more or less all right to me - indeed, there is only one way you can define such $f = \{\} \times S = \{\}$.

The thing I can't get is why $\{\}$ isn't a terminal object? By the same logic, there is only one way to construct $g: S \mapsto \{\} = S \times \{\} = \{\}$.

What am I missing?

P.S. I know that any singleton set is a terminal object in the $\mathcal{SET}$. The question is not about it.

$\endgroup$
5
$\begingroup$

The point is that the subset $f$ of the cartesian product $A\times B$ (for an application $f:A\to B$) must check a property beginning with $$\forall a\in A,\exists b\in B \,\mathrm{s.t.}\dots,$$ so if $A\neq\varnothing$ and $B=\varnothing$, you can't construct a such subset since for a $a$ in $A$, we would have a $b\in\varnothing$, which is impossible.

$\endgroup$
5
$\begingroup$

If $A\neq\varnothing$ then no function $A\to\varnothing$ exists.

This is because for $a\in A$ there is no $f(a)\in\varnothing$.

So $\varnothing$ is not terminal in the category of sets.

$\endgroup$
3
$\begingroup$

The empty "function" $g$ you've constructed isn't a function $S \rightarrow \emptyset$ unless $S$ is empty. If $S$ is not empty then there is an element $x \in S$ which doesn't have an output so $g$ is not a function.

Therefore there are no functions from $S$ into the empty set, rather than a unique function.

$\endgroup$

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.