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The book Theory of Functions of a Real Variable by I. P. Natanson, proves that a denumerable or finite union of pairwise disjoint sets of cardinality $c$ has cardinality $c$.

The proofs given in the book are fairly easy, using the axiom of choice. But then it is left as an exercise a generalization of the above theorems for not necessarily pairwise disjoint sets.

I have searched all over the web, but all the proofs found so far are for pairwise disjoint sets.

The proof given in the book relies on the fact that each set $A_k$ can be matched with a set like $[a_{k-1},a_k)$, with $a_k$ point of a partition of $[0,1]$, such that $\bigcup_{k=1}^{\infty\lor n}A_k=[0,1)$. So it is easy to guess the bijective function, but I can't generalize this proof.

Any help is highly appreciated. Thanks and regards.

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HINT: Let $B_1=A_1$. For $n\ge 2$ let $B_n=A_n\setminus\bigcup_{k=1}^{n-1}A_k$. Then $\bigcup_nB_n=\bigcup_nA_n$, the sets $B_n$ are pairwise disjoint, each has cardinality at most $\mathfrak{c}$, and $B_1$ has cardinality $\mathfrak{c}$.

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    $\begingroup$ @CABJ: $\left|\bigcup_nA_n\right|\ge\left|A_1\right|$. $\endgroup$ – Brian M. Scott Mar 9 '15 at 4:10
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    $\begingroup$ @CABJ: By hypothesis $|A_1|=|Bbb R|$, and $A_1\subseteq\bigcup_nA_n$, so you have a bijection from $\Bbb R$ to a subset of $\bigcup_nA_n$; that’s enough to tell you that $\mathfrak{c}=|\Bbb R|\le\left|\bigcup_nA_n\right|$, and you already have the other inequality. $\endgroup$ – Brian M. Scott Mar 9 '15 at 4:25
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    $\begingroup$ @CABJ: That’s okay. To prove that, use the hint in my answer: replace $\bigcup_nA_n$ with $\bigcup_nB_n$, which is equal to it but a union of pairwise disjoint sets, and apply the result that you already know to the sets $B_n$. $\endgroup$ – Brian M. Scott Mar 9 '15 at 4:37
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    $\begingroup$ @CABJ: The same method as in the theorem that you know will work as long as each of the sets has cardinality at most $\mathfrak{c}$. You won’t be able to get a bijection from $B_n$ to $[a_{n-1},a_n)$, but you will have an injection, which will give you an injection from $\bigcup_nB_n$ to $\Bbb R$; that’s good enough to give you the inequality that you want. $\endgroup$ – Brian M. Scott Mar 9 '15 at 4:53
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    $\begingroup$ @CABJ: You’re welcome! Glad we got it cleared up. $\endgroup$ – Brian M. Scott Mar 9 '15 at 4:58

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