Exercise 3.6.2 (Empty set has cardinality zero)

Proof.

We first show that if a set $X$ has cardinality $0$, then $X=\varnothing$.

Suppose $X$ has cardinality $0$, that is, $\#(X)=0$. Then by Definition 3.6.1 and Definition 3.6.5, there exists a bijection

But $\{i\in N:1\le i\le 0\}=\varnothing$ because we can’t have both $i\ge 1$ and $i\le 0$. Thus, $f:X\to \varnothing$.

Since $f$ is bijective, we have that it is injective (one-to-one) and surjective (onto). Since $f$ is surjective, we have that the forward image is equal to the codomain (Tao calls codomain Range), that is, $f(X)=\varnothing$. Since $f$ is injective, we have, by Exercise 3.4.5,

That is, the domain $X$ equals the inverse image $f^{-1}(\varnothing )$, which, by definition, equals $\{x\in X:f(x)\in \varnothing \}$. But this set must be empty since if it wasn’t, there would exist some object $y\in f^{-1}(\varnothing )$ which would imply $f(y)\in \varnothing$, which is impossible since $\varnothing$ has no elements. Hence, $f^{-1}(\varnothing )=\varnothing$. But we had that $X=f^{-1}(\varnothing )$. Thus, we must have $X=\varnothing$, as was to be shown.

We now show that if $X$ is empty, then it has cardinality $0$.

Suppose $X=\varnothing$. Let $Y$ be the singleton set $\{y\}$; the existence of such a set is guaranteed by the Singleton sets and pair sets Axiom. Then we have $\#(Y)=1$ (this is easy to show). By Lemma 3.6.9, we have $\#(Y-\{y\})=0$. But $Y-\{y\}=\varnothing$, implying $\#(\varnothing )=0⟹\#(X)=0$, as was to be shown.