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Let $A_1, A_2, A_3,\ldots$ be pairwise disjoint denumerable sets.

Prove that $$\bigcup_{i=1}^\infty A_i\text{ is denumerable.}$$

This question comes from exercise $10.11$ of Mathematical Proofs: A Transition to Advanced Mathematics (Second Edition) by chartrand, polimeni, and zhang. I'm using this book for self study.

I used induction in this proof attempt. Sorry for the messy handwriting.

I figured that the natural numbers were equivalent to integers from 1 to infinity, but a buddy pointed out that infinity was not an element of the natural numbers, so induction might not be possible.

So, was it possible to use induction on this problem? Or should I just redo the proof by constructing a table of the denumerable sets and coming up with some diagonal arrows to order them like my book suggests.

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Hockeyfan19
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  • Can you self-study? I mean, grab a math(s) book from somewhere and just start reading? – Mr Pie Jun 15 '18 at 03:21
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    Your buddy is correct. Induction can only be used to prove that $\bigcup_{i=1}^n A_i$ is denumerable for each $n\in \mathbb N$. – K.Power Jun 15 '18 at 03:22
  • I wonder if the book wrote the problem this way to specifically prevent induction proofs.. – Hockeyfan19 Jun 15 '18 at 03:33
  • I could swear that someone had asked recently if they can use induction to prove that a countable union of sets is countable. – Asaf Karagila Jun 15 '18 at 06:54
  • Aha!!! https://math.stackexchange.com/questions/2803257/why-cannot-induction-be-employed-to-show-that-a-countably-infinite-collection-of – Asaf Karagila Jun 15 '18 at 06:57

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You can prove by induction on $k$ that $\bigcup_{i=1}^k A_i$ is denumerable, but that doesn't cover the infinite case, so you need a different strategy. In fact, you need the axiom of countable choice for this. For each $i$, choose one injection $f_i$ from $A_i$ to $\mathbb{N}$. Map each element $x$ of your union to $(i,\,f_i(x))$, with $i$ minimised so that $x\in A_i$. This injects the union into $\mathbb{N}^2$, which is famously countable.

J.G.
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