Definition of finite vs definition of countable set (using Surjective function) So, as can be seen here,
"A non-empty set X is countable if and only if there exists a surjective function f from ℕ onto X
I agree with that. However, in Rosenthal's book on probability theory, in the mathematical appendix, he defines a finite set as
A set $\Omega$ is finite if for some $n\in N$ and some function $f:\mathbb{N} \rightarrow \Omega$, we have $f(\{1,2,3,\dots, n\}) \supseteq \Omega$
Why does he use $\supseteq$? Since $\Omega$ is the codomain of the function, wouldn't the function applied to anything have to be a subset (or equal to, if surjective), the codomain, $\Omega$?
I mean, I guess the $\supseteq$ really is an $=$ since every finite set is countable, and the definition of countable uses equality. However, why use $\supseteq$ in the definition for finite? What benefit does it provide?
 A: The set formed by $f(1,\dots,n) = \{f(1), f(2), \dots, f(n)\}$ is (obviously) a finite set. If $\Omega \subset f(1,\dots,n)$, then it is also finite. But as you have already mentioned, a $=$ would also be appropriate, since all $f(i)$s are in the co-domain.
I guess the usage of $\subset$ just wants to emphasize on the fact, that it would usually be the other way round.
A: It appears to be a (confusing) attempt to economize.
In set theory, contrary to many other areas of mathematics, it is somewhat common to view equality not as a primitive concept, but as a defined notion: Two sets are equal if and only if they have the same elements -- or, in symbols,
$$ A=B \quad\text{means} A\subseteq B \land A\supseteq B $$
Therefore, saying
$$ f:\mathbb N\to \Omega \;\land\; f(1,2,3,…,n)=\Omega $$
would be the same as saying
$$ f:\mathbb N\to \Omega \;\land\; f(1,2,3,…,n)\supseteq\Omega \;\land\; f(1,2,3,…,n)\subseteq\Omega$$
But the last condition is already implicit in $f:\mathbb N\to\Omega$, so it seems a waste (at least to that author) to ask you to check it twice.
The concept being defined is the same as with $=$, though.
