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In a previous, and quite popular, question it was discussed about whether or not $\pi$ contains all finite number combinations.

Let us assume for a moment that $\pi$ does in fact contain all finite combinations of numbers. What prevents $\pi$ from also containing all infinite sets?

It seems that at some point one would also see the first couple digits of, for example, $e$ (2.71828). But why does it need to stop there, couldn't it contain a bunch of digits of $e$? Perhaps even an infinite number of digits of $e$?

My understanding is that $e$ could also be replaced by $\sqrt2$ or any other irrational number, so long as that irrational number contained all finite sets of number combinations. Which might imply that somewhere along the way, $e$ contains a number of digits of $\pi$. Implying this ridiculous situation where within $\pi$ we see $e$, and then within $e$ we again begin to see $\pi$ again. Then all the universe collapses into a singularity. Or maybe someone can just explain why one infinite sequence can't contain another infinite sequence, and perhaps why we have not defined some type of super-infinity that can.

To reiterate the primary question: What prevents $\pi$, or other infinite irrational number that contains all finite sets of numbers, from also containing all infinite sets?

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2 Answers 2

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If the decimal expansion of $\pi$ contained the string of digits '$000\ldots$' then it only has a finite number of non-zero digits (some subset of the digits before the string of $0$s starts) and so $\pi$ is a rational number. Contradiction.

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  • $\begingroup$ That makes a lot of sense. So clear, I almost feel silly for not having realized it. $\endgroup$
    – mwjohnson
    Commented Sep 11, 2013 at 1:35
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    $\begingroup$ The point is that infinite strings don't have an end (on the right as we write them) so there are very few infinite strings actually contained in $\pi$ - only those of the form $[10^n\cdot\pi]$ for $n\geq -1$ where $[\bullet]$ is the 'fractional part' function. $\endgroup$
    – Dan Rust
    Commented Sep 11, 2013 at 1:38
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    $\begingroup$ Rationality and irrationality aside, if it contains 000000... it certainly cannot also contain 111111.... $\endgroup$ Commented Sep 11, 2013 at 2:10
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    $\begingroup$ @NateEldredge Yeah there's a bunch of natural contradictions you can reach. $\endgroup$
    – Dan Rust
    Commented Sep 11, 2013 at 2:15
  • $\begingroup$ Well, you should carefully distinguish between "set" and "sequence". It does contain many different sequences: for instance, 1415926..., 415926..., 15926..., etc, where each one is some "tail" of all the ones before it. But it cannot contain both of two sequences where neither is a tail of the other. $\endgroup$ Commented Aug 6, 2018 at 16:26
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There are an uncountable number of infinite digit strings, and $\pi$ contains only countably many infinite digit strings-namely all the ones that are the digits after some point in the expansion. It does contain infinite strings, just not most of them.

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    $\begingroup$ This is a stronger answer because it also disposes of a lot of variations such as "what if we only want each infinite sequence of digits to appear at some set of positions that form an arithmetic progression?" $\endgroup$ Commented Sep 11, 2013 at 13:05
  • $\begingroup$ @user500668: that depends what you mean by includes. You could start after the fifth decimal place-would you say it includes that sequence? You could take every other term. You could do a lot of things. $\endgroup$ Commented Aug 6, 2018 at 16:26

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