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I came up with this theorem since the proof of $I$ is infinite, $A_k$ is countably infinite, and $A_i$ is countable for all $i \neq k$. Is $\prod\limits_{i\in I}A_i$ countable? is realated to the rearrangement of $A_i$.

While I'm quite sure about the bijection, I'm not sure if my proof for its injection and surjection contains any error. I'm very happy If you can verify that part. Thank you so much!

The Cartesian product of a family $(A_i\mid i\in I)$ is defined as $$\prod\limits_{i\in I}A_i=\{f:I\to\bigcup A_i\mid f(i)\in A_i \text{ for all } i \in I\}$$

Theorem: Let $h:I \to I$ be a bijection. Then $\prod\limits_{i\in I}A_i$ and $\prod\limits_{i\in I}A_{h(i)}$ are equinumerous.

Proof: Consider

$$\begin {array}{l|rcl} G : & \prod\limits_{i\in I}A_i & \longrightarrow & \prod\limits_{i\in I}A_{h(i)} \\ & \phi & \longmapsto & G(\phi) \end{array}$$

where $\phi:I \to \bigcup_{i\in I}A_i$ such that $\phi(i) \in A_i$ for all $i \in I$.

$G(\phi)$ is defined by $G(\phi)(i)=\phi \circ h(i)$ for all $i \in I$.

  1. $G(\phi) \in \prod\limits_{i\in I}A_{h(i)}$

Because $h:I \to I$ is bijective, $h(i) \in I$ for all $i \in I$. Thus $\phi \circ h(i) \in A_{h(i)}$ since $\phi \in \prod\limits_{i\in I}A_i$. It follows that $G(\phi)(i) \in A_{h(i)}$ for all $i \in I$. Hence $G(\phi) \in \prod\limits_{i\in I}A_{h(i)}$.

  1. $G$ is surjective

For $\psi \in \prod\limits_{i\in I}A_{h(i)}$, define $\phi$ by $\phi(i)=\psi \circ h^{-1}(i)$ where $h^{-1}$ is the inverse mapping of $h$.

Then $\psi:I \to \bigcup_{i \in I}A_{h(i)}$ such that $\psi(i) \in A_{h(i)}$ for all $i \in I$.

By definition, $\phi(i)=\psi \circ h^{-1}(i) \in A_{h \circ h^{-1}(i)}=A_i$ for all $i \in I$. Thus $\phi \in \prod\limits_{i\in I}A_{i}$.

By definition, $G(\phi)(i) = \phi \circ h(i) = (\psi \circ h^{-1}) \circ h(i) = \psi \circ (h^{-1} \circ h) (i) = \psi(i)$ for all $i \in I$.

Thus $G(\phi)=\psi$. Hence $G$ is surjective.

  1. $G$ is injective

For $\phi_1,\phi_2 \in \prod\limits_{i\in I}A_i$ and $G(\phi_1)=G(\phi_2)$. Then $G(\phi_1)(i)=G(\phi_2)(i)$ for all $i \in I$, then $\phi_1 \circ h(i) = \phi_2 \circ h(i)$ for all $i \in I$.

Since $h:I \to I$ is bijective, $\phi_1 (i) =\phi_1 \circ h(i')$ and $\phi_2 (i) =\phi_2 \circ h(i')$ for a unique $i' \in I$. Since $\phi_1 \circ h(i') = \phi_2 \circ h(i')$ for all $i' \in I$, $\phi_1 (i) = \phi_2 (i)$ for all $i \in I$.

Thus $\phi_1= \phi_2$. Hence $G$ is injective.

Akira
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1 Answers1

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I agree with the proof in essence.

If you're being this detailed, also check the well-definedness of your maps: check that $G(\phi) \in \prod_{i \in I} A_{h(i)}$ (using the definition of the Cartesian product that you also gave, please do use a capital C in honour of Descartes...) and also in the proof of surjectivity that the $\phi$ you define is in $\prod_{i \in I} A_i$. It's a minor quibble though.

Also you could spell out why $\phi_1(h(i)) = \phi_2(h(i))$ for all $i$ implies $\phi_1(i) = \phi_2(i)$ for all $i$ (namely because all $i \in I$ are of the form $h(i')$ for some $i' \in I$ etc.). I'd say if you're going detailed (i.e. all the gory details) also do these details.

Henno Brandsma
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  • I have added all the detail you suggested :) Please have a look at and check these added parts! Thank you so much! – Akira Sep 01 '18 at 08:52
  • @LeAnhDung Looks OK. I'd write it up a bit differently but that's more of a style matter. – Henno Brandsma Sep 01 '18 at 08:58
  • Many thanks for you ^_^ – Akira Sep 01 '18 at 09:00
  • Hi @Henno I have asked a question at https://math.stackexchange.com/questions/2907522/prove-this-version-of-recursion-theorem-from-textbook-introduction-to-set-theory for several days, but have not received any answer. Could you please help me check it out? – Akira Sep 11 '18 at 23:50