How do I prove that two empty sets are equivalent? I need to prove that two empty sets have the same cardinality,
I know that proving equivalence is done by defining an onto and one-to-one function from one group to the other, but how can I define a function from an empty set to an empty set?
 A: Let $A=B=\varnothing$. A function from $A$ to $B$ is a subset $f$ of $A\times B$ with the property that if $\langle a,b_1\rangle$ and $\langle a,b_2\rangle\in f$, then $b_1=b_2$. The only subset of $A\times B=\varnothing\times\varnothing=\varnothing$ is $\varnothing$. Does it have the function property? Yes: it contains no ordered pairs at all, so the hypothesis 

if $\langle a,b_1\rangle$ and $\langle a,b_2\rangle\in f$

is never satisfied, and the property

if $\langle a,b_1\rangle$ and $\langle a,b_2\rangle\in f$, then $b_1=b_2$

never actually requires anything to be true $-$ it holds vacuously.
Thus, $\varnothing$ is a function from $\varnothing$ to $\varnothing$. Is it a bijection? Is it true that for each $a\in A=\varnothing$ there is a unique $b\in B=\varnothing$ such that $\langle a,b\rangle\in f$? Again the answer is yes, vacuously: there’s no $a\in A=\varnothing$ at all, so the requirement is never actually invoked.
However, all of that work is completely unnecessary, because it’s always true that if $A=B$, then $|A|=|B|$. And if $A$ and $B$ are both empty, then they have exactly the same members, so they are the same set, a fact that I implicitly used above when I wrote $A=B=\varnothing$.
A: Hint: Recall that there are no elements in an empty set. Show that if $A$ and $B$ are empty sets then for every $x\in A$ we have $x\in B$, and vice versa. Therefore $A\subseteq B$ and $B\subseteq A$ and so $A=B$.
If two sets are equal then certainly they have the same cardinality.
A: Hint:
Use the definition of the (an) empty set - a set with no elements - to show, it is vacuously true that for all $x$ in the domain, $\;x \in \emptyset_1 \implies x \in \emptyset_2,\;$ so $$\emptyset_1 \subseteq \emptyset_2$$ and do similarly to show that it is vacuously true that $$\emptyset_2 \subseteq \emptyset_1$$
Hence, $\emptyset_1 = \emptyset_2$.
Hence, $|\emptyset_1| = |\emptyset_2|$
A: The extension and specification axioms implies that there is only one set with no elements.
