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I come up with this theorem when I prove: If $A$ is uncountable and $B$ in non-empty, then $A \times B$ is uncountable. I found that proving this is not so hard by creating a bijection $f:A \times \{b\} \to A$ where $b \in B$ and notice that $A \times \{b\} \subseteq A \times B$. But I'm not sure if my proof for the generalization contains any error.

Please help me check the part that I define mapping $G$!


Lemma: If $f:A \to B$ is surjective and $A$ is countable, then $B$ is countable.

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 $(A_i \mid i \in I)$ be the family of non-empty indexed sets where $A_k$ is uncountable for some $k \in I$. Then $\prod\limits_{i\in I}A_i$ is uncountable.

We define a mapping $G:\prod\limits_{i\in I}A_i \to A_k$ by $G(f)=f(k) \in A_k$ for all $f \in \prod\limits_{i\in I}A_i$. Then $G$ is clearly surjective. Assume the contrary that $\prod\limits_{i\in I}A_i$ is countable. Then, by Lemma, $A_k$ is countable, which is a contradiction. Hence $\prod\limits_{i\in I}A_i$ is uncountable.

Update: On the basis of Kavi Rama Murthy's and Saucy O'Path's comments, I added the proof that $G$ is surjective here.

By Axiom of Choice, there exists a mapping $f:I\to\bigcup A_i$ such that $f(i)\in A_i$ for all $i \in I$. For any $a \in A_k$, we define a mapping $f_a$ by $f_a(i)=f(i)$ for all $i \neq k$ and $f_a(k)=a$. Then $f_a$ is clearly in $\prod\limits_{i\in I}A_i$ and $G(f_a)=a$. Hence $G$ is surjective.

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  • $\begingroup$ Perfect! Your proof is valid (assuming Axiom Of Choice). However, you have assume that $A_i$'s are all non-empty. If some $A_i$ with $i\neq k$ is empty then the product will become empty! $\endgroup$ – Kavi Rama Murthy Aug 31 '18 at 10:09
  • $\begingroup$ @KaviRamaMurthy I have added that $A_i$ is nonempty for all $i \in I$ :) Of course, we must appeal to Axiom of Choice when dealing with $I$ in general. $\endgroup$ – Crush_on_You Aug 31 '18 at 10:16
  • $\begingroup$ @KaviRamaMurthy I have asked a question at math.stackexchange.com/questions/2907637/… for several days, but have not received any answer. Although I have some questions that are not answered such as this one. I'm very in need of getting an answer for this question. Could you please help me check it out? $\endgroup$ – Crush_on_You Sep 11 '18 at 23:49
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Claiming that $G$ is surjective is a slightly delicate matter, because it requires you to prove that for all $u\in A_k$ there is a function $f:I\to \bigcup_{i\in I} A_i$ such that $f(i)\in A_i$ for all $i\in I$ and $f(k)=u$. This is certainly false if $A_i=\emptyset$ for some $i\ne k$ (and the entire thing is false in that instance). Giving a proper argument for the case where $A_i\ne \emptyset$ for all $i$ is tantamount to the axiom of choice.

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  • $\begingroup$ I have added that $A_i$ is nonempty for all $i \in I$ :) Of course, we must appeal to Axiom of Choice when dealing with $I$ in general. $\endgroup$ – Crush_on_You Aug 31 '18 at 10:17
  • $\begingroup$ Please check my argument that $G$ is surjective! By Axiom of Choice, there exists a mapping $f:I\to\bigcup A_i$ such that $f(i)\in A_i$ for all $i \in I$. For any $a \in A_k$, we define a mapping $f_a$ by $f_a(i)=f(i)$ for all $i \neq k$ and $f_a(k)=a$. Then $f_a$ is clearly in $\prod\limits_{i\in I}A_i$ and $G(f_a)=a$. Hence $G$ is surjective. $\endgroup$ – Crush_on_You Aug 31 '18 at 10:22
  • $\begingroup$ It is perfeclty fine. $\endgroup$ – Saucy O'Path Aug 31 '18 at 10:27
  • $\begingroup$ Thank you so much ^^ $\endgroup$ – Crush_on_You Aug 31 '18 at 10:27
  • $\begingroup$ Hi @Saucy I have asked a question at math.stackexchange.com/questions/2909659/… for several days, but have not received any answer. Could you please help me check it out? $\endgroup$ – Crush_on_You Sep 11 '18 at 23:51

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