Alternate (intuitive) definition of Cardinality for Infinite Sets For cardinality of infinite sets, the definition commonly used in mathematics is in terms of injectivity and surjectivity. If a set has a one-to-one correspondence with another, both sets have the same cardinality. If not, one is "bigger" than the other. In the case of finite sets, this definition reduces to comparing the amount of elements in the sets.
A 'surprising' result of this definition is that the cardinality of the natural numbers and the odd numbers is the same. It's surprising because it goes against 'intuition'.
But if it goes against 'intuition', then there must be an alternative definition for infinite sets, one that is more 'intuitive'. I believe this alternative definition is based on the limit of the size of the finite set, as its size goes to infinity.
With this definition, the ratio of the size of the natural and odd numbers, inside a given range (1 to N, for example), when N goes to infinity, is either exactly 1/2, or a number that approaches 1/2. So by taking the limit of their size, as N goes to infinity, we find that the alternative-cardinality of the odds is exactly 1/2 that of the naturals.
My questions are: is this alternate definition recognized by mathematicians? Is it used somewhere?
P.S: If I say the size of the naturals is twice the size of the odds, would it really be incorrect? I feel that the concept of size, for infinite sets, is subjective and there is more than one acceptable answer.
 A: One place where your construction is of fundamental use is geometric group theory.  You may have heard of the Banach-Tarski paradox: One can disassemble a ball in three dimensional space, into a finite number of pieces, and then move the pieces (through rotations and translations) and reassemble them as two balls, each the size of the original.
One of the key ideas is to consider a pair of rotations of the ball, $A,B$ such that each sequence of rotations (e.g. word in the letters $A,B$) results in a different rotation.  The point here is that we can split all words in the letters $A,B$ into those that begin $A$  and those that begin $B$.  Denote these sets of words $W_A,W_B$ respectively.  Then removing the first letter from each word in $W_A$, we get an entire copy of all words.  Similarly removing the first letter from each word in $W_B$ we get all words.
So far this is a very vague explanation, but let's look in more detail at why this does not work  for the punctured disk, and we will see how your idea is key to this.
Suppose we can partition the integers into a finite number of sets $A_1,\cdots,A_n,B_1,\cdots,B_m$ such that that we have integers $a_1,\cdots,a_n,b_1,\cdots,b_m$ and the transled sets $a_1+A_1, \cdots,a_n+A_n$ form a disjoint partition of the integers, as do the translated sets, $b_1+B_1, \cdots,b_m+B_m$.
Then let $\theta$ be an irrational number, so the $\{e^{2\pi ir\theta}|\,r\in \mathbb{Z}\}$ form a subgroup (isomorphic to the integers) of the unit complex numbers under multiplication. Let $X$ be a complete system of coset representatives for this subgroup (we need the axiom of choice to guarantee this exists).  Let $\hat{X}$ denote all complex numbers of the form $\lambda x'$ with $\lambda\in (0,1]$ and $x'\in X$.
Then every point on the punctured disk $\{z\in\mathbb{C}|\,\,0<|z|\leq 1\}$ can be uniquely written in the form $e^{2\pi ir\theta} x$, for some integer $r$ and $x\in \hat{X}$.
Then we can partition the punctured disk into pieces: $$D_j=\{e^{2\pi ir\theta} x|\,\, r\in A_j, x\in \hat{X}\},\,\,{\rm and}\,\,E_k=\{e^{2\pi ir\theta} x|\,\, r\in B_k, x\in \hat{X}\}.$$
Then rotating each $D_j$ by multiplication by $e^{2\pi ia_j\theta}$ and reassembling we get a complete copy of the punctured disk.  Similarly translating all the $E_k$ by adding say $3$ to each complex number in each $E_k$, and then rotating each $E_k$ by $e^{2\pi ib_k\theta}$, we get another copy of the punctured disk.
The point is that if we could split up the integers into the $A_j,B_k$ as above, then we could duplicate the punctured disk, the same way that we can duplicate a $3$ dimensional ball.  However it turns out that area behaves differently to volume, and cannot be doubled in this way.  In particular, the integers behave differently to words on two letters, and cannot be duplicated the same way.
The proof that the decomposition into $A_j,B_k$ is impossible is basically just your idea of "alternative cardinality".  Firstly, as we are now working with the integers, the ranges need to be $-N,\cdots,N$.  Also, as I mentioned in comments, the ratios will not always converge.  However the ratios  all lie in $[0,1]$ so by the Bolzano–Weierstrass theorem, they have convergent subsequences.  We use the notion of an ultralimit, to pick out a particular limit.
It turns out that with these modifications to your construction $|S_1 \cup S_2|=|S_1|+|S_2|$ for disjoint subsets $S_1,S_2$ of the integers, $|\mathbb{Z}|=1$ and $|S+k|=|S|$ for any subset $S$ of the integers, and integer $k$.
Then: \begin{eqnarray*}
1&=&|\mathbb{Z}|\\&=&\sum_{j=1}^n|A_j|+\sum_{k=1}^m|B_k|\\
&=&\sum_{j=1}^n|a_j+A_j|+\sum_{k=1}^m|b_k+B_k|\\
&=&|\mathbb{Z}|+|\mathbb{Z}|=1+1=2,
\end{eqnarray*}
giving the desired contradiction.
Here are some wiki links on the various ideas discussed:
Banach-Tarski paradox
Ultralimit
F$\emptyset$lner Sequence (these are a generalisation of range $-N,\cdots,N$)
Amenable group
