On page 25 of Principles of Mathematical Analysis (ed. 3) by Rudin, there is the definition (excluding the irrelevant parts for this question):
Definition 2.4: For any positive integer $n$, let $J_n$ be the set whose elements are the integers $1,2,...,n$; let $J$ be the set consisting of all positive integers. For any set $A$, we say:
(a) $A$ is finite if $A \thicksim J_n$ for some $n$ (the empty set is also considered to be finite).
I am unsure why the empty set is considered to be finite. Given the defintion, for the empty set to be finite, then $\emptyset \thicksim J_n$ for some $n$.
The issue is for equivalence, there has to exist a one-to-one mapping of $\emptyset$ onto $J_n$. $\emptyset$ has no elements to correspond and since the definition requires $n \in J$, $J_n$ will always have at least one element.
Thus, my question: Why is the empty set considered finite?
That definition is a (rare) example of Rudin doing things inefficiently. He could have defined $J_n$ for each non-negative integer $n$ to be the set of non-negative integers less than $n$, so that $J_0=\varnothing$, $J_1=\{0\}$, $J_2=\{0,1\}$, etc. Then he could have defined a set $A$ to be finite if and only if $A\sim J_n$ for some $n\in\Bbb N$ (where $\Bbb N$ includes $0$). This is essentially the usual set-theoretic definition stripped of some set-theoretic detail that would be out of place here.