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Is the set of all digits in a real number countable/listable? It seems to me that it is, based merely on my layman's reading of popular books about set theory. In other words, it seems that one can map each position in a real number to one of the natural numbers.

But as usually happens with math, I am probably missing something.

Thanks for any help.

Question edited based on the answer from fleablood. I think he has properly paraphrased my confusion on this topic.

"Perhaps you are thinking that if sequence that makes up the real number is countable, that that should mean the number of real numbers (or equivalently the number of possible countable sequences) should also be countable. But you know that there is a rule written in stone 'the Real Numbers are Uncountable' and you are wondering where your error is.

It is not the case that the number of countable sequences should be countable. Just there are an infinite number of finite numbers, there are an uncountable number of countable sequences."

Yes, that was my thinking almost exactly. I know the real numbers have been proven to be uncountable, but I have trouble understanding how.

For example, the set of digits in real number r1 is countable. The set of digits in real number r2 is also countable. If you add them, r1 + r2, you get real number r3 whose set of digits is also countable. I imagine a never-ending series such as r1 + r2 + r3 ... rN. If that sum is always a real number with a countable set of digits I don't understand where along the line the result is an uncountable set.

My guess is that I am begging the question somehow (r1 + r2 + r3 ... rN doesn't say anything about the size of the set being used for the sum) but it still nags me.

Is there a layman's version of the proof for "there are an uncountable number of countable sequences"?

Thanks again!

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  • $\begingroup$ To your last question, yes, diagonalization is super easy. $\endgroup$ – Asaf Karagila May 24 '18 at 16:28
  • $\begingroup$ Thanks Asaf. I know about diagonalization, as it pertains to "entire" real numbers (that there is always a way to generate a real number not in the list). I guess the same can be applied to the set of digits within a real number? I'll try to Google that to see if I can find a simple explanation. $\endgroup$ – jrdevdba May 24 '18 at 19:57
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The set of all decimal digits of a real number is certainly countable, since they can be put into 1-to-1 correspondence with the positive integers - we can define the first digit, the second digit, the third digit and so on.

Note that this does not mean that every real number is computable - even though the digits of a number are countable, there is not necessarily an algorithm for calculating them.

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Well, the digits are just $\{0,1,2,3,4,5,6,7,8,9\}$ so there are at most ten. I think what you meant is say is the is the sequence $\{a_i\}$ where $a_i$ is the $i$-th digit countable.

The answer is ... Yes.... They are indexed in a list one after another and are therefore in 1-1 corespondence with $\mathbb N$ with $a_i \leftrightarrow i$. You can .... count them... "after the $1,257$th digit which is $5$, is the $1,258$th digit which is $2$ and then the $1,259$th digit is...."

This is absolutely true and you thoughts are correct.

Perhaps you are thinking that if sequence that makes up the real number is countable, that that should mean the number of real numbers (or equivalently the number of possible countable sequences) should also be countable. But you know that there is a rule written in stone "the Real Numbers are Uncountable" and you are wondering where your error is.

It is not the case that the number of countable sequences should be countable. Just there are an infinite number of finite numbers, there are an uncountable number of countable sequences. And the reasons are kind of similar.

If $n$ was the last finite number then we get a contradiction when we condiser $n+1$ which is also a finite number. So there can be no last finite number. So the number of finite numbers is infinite.

If $\{s_1, s_2, ........\}$ were a countable list of all countable sequences. then the sequence $t$ where $t_i \ne s_{i_i}$ will not be on the list. But it is a countable sequence. So then number of countable sequences is uncountable.

....

Oh, and BTW, yes the number of countable sequences of the ten digits is the same as the number of real numbers and we can put the countable sequences into a decimal representation and do that for every real number so there will be a 1-1 corespondence between every countable sequence of digits and real numbers. But I'm not going to prove or argue that here.

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Say $x$ is a real number. Then the set $D_x$ of digits of $x$ is (at least for a 10-adic representation of $x$) a subset of the finite set $\{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$ and therefore $D_x$ itself must be finite and thus also countable.

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