The infinite sequence of the digits of pi

[jbriggs444]

But the point is this: As you go from countable to uncountable, the value of the string changes from exactly equal to Pi, to "won't converge".

Even if the construction that you describe were a construction, it does not go from countable to uncountable. A countable collection of countable collections is countable.

 

[Algr]

But doesn't Cantor's Diagonal show that the decimals are uncountable? Putting Pi in that format would show that Pi's digits are uncountable too.

 

[jbriggs444]

But doesn't Cantor's Diagonal show that the decimals are uncountable? Putting Pi in that format would show that Pi's digits are uncountable too.

It shows that any countable list of countable decimal strings must be incomplete. Every such list must miss at least one decimal string.

It does not show anything about the cardinality of the set of digit positions in a decimal string. It does serve to show that regardless of the cardinality of the set of digit positions, the cardinality of the set of all strings with decimal digits in those positions must exceed that.

If you decide to consider decimal "strings" with $\aleph_1$ many digit positions then you will have a set of decimal "strings" with a cardinality in excess of $\aleph_1$.

If you decide to consider decimal strings with 2 digit positions then you will have a set of decimal strings with a cardinality in excess of 2. Indeed, 100 is greater than 2.

It is not clear what "putting $\pi$ in that format" even means.

Edit: Some people glance over Cantor's diagonal argument and think that it is an algorithm for creating an uncountable list by iteratively appending more and more numbers at the end of a countable list. That is not how the argument works. It works as @FactChecker outlines below.

 

[FactChecker]

But doesn't Cantor's Diagonal show that the decimals are uncountable? Putting Pi in that format would show that Pi's digits are uncountable too.

The diagonal proof constructs one number that is not on the proposed countable list. That number is still represented by a countable number of digits. It is not the countable digits that keeps the list countable. The thing that keeps the list countable is that you can count the members on the list.

PS. If you look at a Cantor diagonal proof that uses the decimal system to represent numbers, you can see that there are enormously more ways to construct numbers not on the list than there are numbers on the list.

 

[fresh_42]

The sequence of decimal digits of any number written in decimal digits is countable per construction. Just count them, one by one. This will be short in the case you count $0.5$ and will take a bit longer in the case of $\pi.$

However, it could be interesting to discuss the term pseudo-randomness or whether $\pi$ is or is not pseudo-random, but this would be another subject. The digits themselves are only one - and insufficient - way to write $\pi.$ We still do not know enough members of A000796 and its properties. Countability, at least, is certain.

This thread is closed.