Cardinality of a countably infinite set, once partitioned

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.

• 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

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.
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 $0 \leq r <n$.