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What is the cardinality of a countably infinite set, once we have partitioned it into finitely many countably infinite subsets?

My question might appear stupid, but I just can't decide/prove whether the answer is finite or countably infinite.

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Countably infinite. Partitioning it won't change its size. –  Deven Ware Jan 30 '13 at 18:24
    
The basic property of a set is that it is a collection of 'things' (whatever they may be, usually given by some kind of description) so that you can tell whether a given 'thing' is in the set or is not in the set. If you partition the set somehow, it will not change whether or not a given 'thing' is in the set or not and so it will not change the set. –  Barbara Osofsky Jan 30 '13 at 21:32
    
Let me guess what prompted your question. You may be familiar with "clock arithmetic, alias the integers modulo 12. This is a finite set of subsets of integers. $\left\{\ 0\,\ \pm 12\ , \ \pm 24\ ,\ \cdots\right\} $ is an element of the integers modulo 12, it is not an integer. $0$ is a representative of that member of the integers module 12, but so is 12 or 240 or $\cdots$. You must (try to) always distinguish between elements of sets and subsets of sets. Hope this helps. You question does not appear stupid to me. It appears to indicate a trap many of us fall into. –  Barbara Osofsky Jan 30 '13 at 21:53
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2 Answers

The partition is finite, but the set itself is still countably infinite. You can even guarantee, in the case where the partition is arbitrary, that at least one of the parts is countably infinite.

I poured a 750ml bottle of whisky into three glasses. Do I still have 750ml of whisky? Of course. I haven't drank it yet. Even though it's now in three distinct glasses and not in a bottle, 750ml is still 750ml.

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+1, but I’ll bet that you lost some whisky to evaporation. :-) –  Brian M. Scott Jan 30 '13 at 18:41
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@Brian: No, all my chickens are spherical in vacuum. :-) –  Asaf Karagila Jan 30 '13 at 18:48
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Let $S$ be such a countably infinite set. Then now partition this set as $$S_{1} \bigcup \cdots \bigcup S_{n} = S$$ setting that $S_{i}$ is countably infinite. Now you can count its elements in such a way that $$ S_i = \{a_{(i,1)},\dots,a_{(i,k)},\dots \}. $$ In this way, you can count all of them by using sequence. Therefore, the LHS is also countably infinite.

But you obviously know that even without counting them, their cardinalities are the same because they are the same set.

A more explicit bijection can be given: set $f: S \to S_{1} \bigcup \cdots \bigcup S_{n}$ as $$f(s_{j}) = a_{(r,q+1)} $$ where $S = \{ s_{j} \} _{ j \in \mathbb{N} } $ and $j = nq + r$ with $0 \leq r <n $.

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