Here, we will prove that the only set which satisfies the universal property of the empty set is the empty set itself. This will tell us that defining the empty set by this universal property is actually a coherent thing to do, because it’s not ambiguous as a definition.

There are three ways to prove this fact: one way looks at the objects themselves, one way takes a more maps-oriented approach, and one way is sort of a mixture of the two. All of the proofs are enlightening in different ways.

Recall first that the universal property of the empty set is as follows:

The empty set is the unique set \(X\) such that for every set \(A\), there is a unique function from \(X\) to \(A\). (To bring this property in line with our usual definition, we denote that unique set \(X\) by the symbol \(\emptyset\).)

The “objects” way

Suppose we have a set \(X\) which is not empty. Then it has an element, \(x\) say. Now, consider maps from \(X\) to \(\{ 1, 2 \}\).

We will show that there cannot be a unique function from \(X\) to \(\{ 1, 2 \}\). Indeed, suppose \(f: X \to \{ 1, 2 \}\). Then \(f(x) = 1\) or \(f(x) = 2\). But we can now define a new function \(g: X \to \{1,2\}\) which is given by setting \(g(x)\) to be the other one of \(1\) or \(2\) to \(f(x)\), and by letting \(g(y) = f(y)\) for all \(y \not = x\).

This shows that the universal property of the empty set fails for \(X\): we have shown that there is no unique function from \(X\) to the specific set \(\{1,2\}\).

The “maps” ways

We’ll approach this in a slightly sneaky way: we will show that if two sets have the universal property, then there is a bijection between them. noteThe most useful way to think of “bijection” in this context is “function with an inverse”. Once we have this fact, we’re instantly done: the only set which bijects with \(\emptyset\) is \(\emptyset\) itself.

Suppose we have two sets, \(\emptyset\) and \(X\), both of which have the universal property of the empty set. Then, in particular (using the UP of \(\emptyset\)) there is a unique map \(f: \emptyset \to X\), and (using the UP of \(X\)) there is a unique map \(g: X \to \emptyset\). Also there is a unique map \(\mathrm{id}: \emptyset \to \emptyset\). noteWe use “id” for “identity”, because as well as being the empty function, it happens to be the identity on \(\emptyset\).

The maps \(f\) and \(g\) are inverse to each other. Indeed, if we do \(f\) and then \(g\), we obtain a map from \(\emptyset\) (being the domain of \(f\)) to \(\emptyset\) (being the image of \(g\)); but we know there’s a unique map \(\emptyset \to \emptyset\), so we must have the composition \(g \circ f\) being equal to \(\mathrm{id}\).

We’ve checked half of “$f$ and \(g\) are inverse”; we still need to check that \(f \circ g\) is equal to the identity on \(X\). This follows by identical reasoning: there is a unique map \(\mathrm{id}_X : X \to X\) by the fact that \(X\) satisfies the universal property noteAnd we know that this map is the identity, because there’s always an identity function from any set \(Y\) to itself., but \(f \circ g\) is a map from \(X\) to \(X\), so it must be \(\mathrm{id}_X\).

So \(f\) and \(g\) are bijections from \(\emptyset \to X\) and \(X \to \emptyset\) respectively.

The mixture

This time, let us suppose \(X\) is a set which satisfies the universal property of the empty set. Then, in particular, there is a (unique) map \(f: X \to \emptyset\).

If we pick any element \(x \in X\), what is \(f(x)\)? It has to be a member of the empty set \(\emptyset\), because that’s the codomain of \(f\). But there aren’t any members of the empty set!

So there is no such \(f\) after all, and so \(X\) can’t actually satisfy the universal property after all: we have found a set \(Y = \emptyset\) for which there is no map (and hence certainly no unique map) from \(X\) to \(Y\).

This method was a bit of a mixture of the two ways: it shows that a certain map can’t exist if we specify a certain object.