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We know that if we have two sets $A,B$, $A \times B$ is countable. So basically, the cartesian product of two sets is countable. Then, we know $(A \times B) \times C$ is countable, which means $((A \times B) \times C) \times D$ is countable, and so on and so forth for infinity. Why does this fail?

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    $\begingroup$ By doing this you can inductively prove that any finite product of countable sets is countable however you never actually consider a countable (or uncountable) cartesian product of countable sets. $\endgroup$
    – Ben S.
    Aug 30 '21 at 16:24
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    $\begingroup$ The product $\{0,1\} \times \{0,1\} \times \cdots $ is basically the binary representation of the numbers on $[0,1]$ which is uncountable. $\endgroup$
    – copper.hat
    Aug 30 '21 at 16:24
  • $\begingroup$ Ahem... Which of these sets are countable? It is obviously not true that the product of any two sets is countable. $\endgroup$
    – user562983
    Aug 30 '21 at 16:25
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    $\begingroup$ "and so on and so forth for infinity" is hiding all the important details $\endgroup$ Aug 30 '21 at 16:26
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    $\begingroup$ Showing something is true for all finite values of, say $n$, does not mean it is true for an infinite number. $\endgroup$
    – copper.hat
    Aug 30 '21 at 16:26
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$\{0\}$ is a finite set
$\{0,1\}$ is a finite set
$\{0,1,2\}$ is finite set

.....Repeating to infinity gives us that the set of all naturals is a finite set......

All you can prove using induction is that the Cartesian product of a finite number of countable sets is countable

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    $\begingroup$ This is a good example to keep in mind. $\endgroup$
    – Randall
    Aug 30 '21 at 17:14

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