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Is a class of $n< \infty$ infinite sets a finite class?

My Thought

Suppose we have a class $S=\{N_1,N_2,...N_n\}$ where $N$ is a set with infinite members and $n<\infty$

Since $N_i \in S$, $i=\{1,2,...,n\}$ is itself a member of $S$, not its components. By definition, a class is finite if it has a finite $n$ members. Therefore, $S$ is a finite class.

What I was Wondering

Is my reasoning correct?

If not, what is the proper way to look at this?

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You are correct: $S$ is a finite set, with precisely $n$ members (assuming that the sets $N_1,\dots,N_n$ are distinct sets). – Brian M. Scott Jan 30 '13 at 8:54
up vote 1 down vote accepted

Yes, this is correct.

The correct way is to verify that the definition of "finite" holds for $S$. The slightly more "mathematical" definition of finite would be:

$S$ is finite if and only if there exists a bijection $f$ from $S$ onto a set of $\{0,\ldots,n-1\}$, where $n\in\Bbb N$.

We can assume that $N_i\neq N_j$, then the function $f(N_i)=i-1$ is a bijection, as wanted.

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