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I've been reading a basic book on set theory for a while now. And recently ran into an uncountability of $(0, 1)$. I guess like almost everyone who tries to understand what he or she reads I tried to come up with a rule that matches every real number in $(0, 1)$ with a corresponding natural number to see why it is not trivially possible.

Well I came up with something similar to what's described here:

I also recognised that my example doesn't take into account real numbers with infinite decimal representation. But here's the thing. Yes, I cannot write down a natural number that corresponds to $0.3(3)$, but for the same reason I cannot write down the natural number that corresponds to a natural number $33...$. This fact (as far as I know) doesn't make natural numbers uncountable. If we accept $33...$ as a valid representation of the natural number that has all $3$s in it's decimal representation, will it make $(0, 1)$ countable?

Thanks.

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John Doe
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1 Answers1

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If you allow natural numbers to have infinitely many digits, then this collection is not countable anymore.

Yes, if you map a real number to its string of digits (even if you have two possible representations, e.g. $0.4\bar9=0.5$ you can always choose the finite one) then this is a well-defined injection. But you no longer inject the real numbers into a countable set. So you in fact changed the definition of countable.

So no, the real numbers are not countable. But they are of the same cardinality as the cardinality of all infinite sequences of digits, of natural numbers.

Asaf Karagila
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