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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?

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majmun
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  • That symbol does not mean "is a subset of". It means "contains". He is saying a set is finite if it is covered by a finite set of values. In this case, the values are $f(1), f(2), ... f(n)$. – John Douma Nov 09 '15 at 20:30
  • You are correct that if $\Omega$ is the codomain of $f$, then $f(A) \subseteq \Omega$ for any subset $A$ of the function's domain. So $f(A) \supseteq \Omega$ if and only if $f(A) = \Omega$. I agree that it is a somewhat strange way to write it, but I don't have the book so can't see the context. Maybe he wants to emphasize that if you want to check that $f(A) = \Omega$, you only need to verify the containment $f(A) \supseteq \Omega$, since the reverse containment is always true. –  Nov 09 '15 at 20:31

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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.

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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.

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