Why is the empty set finite? 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?
 A: The parenthetical remark is just saying that formally the definition of finite set does not apply to the empty set, but the empty set is taken to be finite by convention. 
If this bothers you, note it is possible to define a set as infinite precisely when there exists a proper subset of it with a bijection to the set. This captures in one go what it means to be infinite.
A: What they are saying here is that the empty set is considered to be finite by convention, or for a different reason. If they were being more precise, they would have said that a set is finite if it is empty or is in bijection with some $J_n$.
Suppose in general that for non-negative integers $n$ we defined $J_n$ to be the set of positive integers no greater than $n$. This certainly matches the definition given for $J_n$ when $n\ge 1$, but what about when $n=0$? Well, there aren't any positive integers no greater than $0$, so $J_0=\emptyset$! Certainly the empty set is in bijection with itself. In that way, we may extend the definition of finite, without relying on a "just roll with it" in the case of the empty set.
A: 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.
A: We know that an empty set is a subset of all sets. If an empty set is infinite, how will it be subset of all sets. So, an empty set is must be finite.
