Definition. Two sets have the same cardinality iff they can be put into one-to-one correspondence; or,
$$A \simeq B \iff |A| = |B| $$
This definition applies to infinite as well as to finite sets.
It follows from the last three definitions that set $A$ has a larger cardinality than set $B$ iff both
- a proper subset of $A$ and the whole of $B$ can be put into one-to-one correspondence
- the whole of $A$ cannot be put into one-to-one correspondence with any
proper subset of $B$.
From: A Crash Course in the Mathematics Of Infinite Sets
Peter Suber, Philosophy Department, Earlham College
The last part specifically refers to proper subsets of $B$. That excludes at least one subset of $B$, namely $B$. So, is it possible that the whole of A cannot be put into one-to-one correspondence with any proper subset of $B$ but may with $B$ itself? Why did he not say "any subset of $B$"?
Best Answer
Any infinite set $B$ has at least a proper subset $C$ such that $|C|=|B|$ (assuming choice, of course, or some weaker axiom thereof).
Since $B$ is infinite, it is not empty. Let $b_0\in B$ and consider $C=B\setminus\{b_0\}$.
Then $|B|=|C|$. Indeed, take a countable subset $Z$ of $B$ (it exists by choice). Then $Z\cup\{b_0\}$ is countable as well, so we can assume $b_0\in Z$. There exists a bijection $f\colon\mathbb{N}\to Z$ such that $f(0)=b_0$. Now consider $F\colon B\to C$ defined by $$ F(x)=\begin{cases} f(n+1) & \text{if $x\in Z$ and $x=f(n)$} \\[4px] x & \text{if $x\notin Z$} \end{cases} $$ It's easy to prove that $F$ is a bijection.
If $|A|=|B|$, then we can use $F$ to provide also a bijection $A\to C$ and $C$ is a proper subset of $B$.