Does Pi contain itself?

Question

Alright, recently there was a question on 9gag whether the digits of $\pi$ may contain $\pi$ itself here's the original. One user had - in my opinion - a really plausible answer:

Here's his answer.

Nevertheless, I would still argue that $\pi$ can "contain" itself. Since $\pi$ is a non-ending sequence of decimal places, consider that the starting index of the repetition can be an infinitely large natural number. We therefore have an infinite non-repeating sequence prior to that index, thus it would still fulfill the requirement of non-repetitiveness in the sense of $ \pi \not\in \{a/b\} \; \forall \; a,b \in \mathbb{N}$.

I wanted to know if this argument is reasonable and what you think about it.

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@5xum As for your first remark:

Let that starting index be $\aleph_0$, the smallest infinite cardinal number.

As for your second remark:

Assume the $(n+1)$-th digit is $\aleph_0$.

Answer 1

Your argument requires further elaboration before it can be considered a "mathematical argument":

• In particular, you need to explain what a "infinitely large natural number" is.
• You then need to explain, in your new mathematical system, what the technical definition of "containing itself" means. In the standard sense, it means that there exists some $n\in\mathbb N$ such that the digits $n+1,n+2,\dots$ are equal to digits $1,2,3,\dots$, respectively. If you allow infinite numbers, then you need to define what the "$n+1$-th digit" is for $\pi$, where $n$ is an infinitely large natural number.

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Oh, and another thing. The answer you link to is not only plausible, it is correct. There is no strict subsequence of digits of $\pi$ which would be equal to the entire sequence of digits of $\pi$, and this is a mathematically provable fact. Calling this fact "plausible" is basically insulting to it.