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There was a discussion about if there can be 2 infinite sequences $a=a_0a_1a_2...$ and $b=b_0b_1b_2...$ over the set $\{0..9\}$ that both appear in one infinite sequences$c=c_0c_1c_2...$

(Actually the discussion was if 2 Infinite sequencescan be found within $\pi$. >_<

The discussion was held by Computer Science students who obviously have no clue of maths. How do I explain them rather easy that it is impossible?

I tried explaining that if you set $c=a_0a_1a_2...b_0b_1b_2$, c is not a sequences where b appears ever. They invented a new theory of sequences where $...b_2b_1b_0a_0a_1a_2...$ is a sequences...

Obviously you can interlace the sequences to $c=a_0b_0a_1b_1...$ but they are convinced that they can construct a sequences which contains a and b as a whole. Any help proving them wrong?

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    Don't worry, it happens to most of us here and there.2012-12-26

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I thnk this is about (digit) sequences. Those are essentially maps from the set $\mathbb N$ to the set $\{0,\ldots,9\}$ of digits. If $b_0b_1\ldots$ appears as a (contiguous) subsequence then there the digit $b_0$ appears at a specific position $n\in\mathbb N$ and has only finitely many digits (namely $n-1$) to its left, hence not another infinite sequence.

They can well invent their "new theories", where we consider maps defined on $\mathbb Z$ and thus have infinite sequences to the left and to the right. Actually, we can also have two infinite sequences following each other if we move from $\mathbb N$ (or $\omega$ as one would rather say in this context) to bigger ordinals, for example $\omega+\omega$ which is just that: two copies of $\mathbb N$ following each other. But both these modifications do not match what is generally defined as "sequence" (and what is used implicitly as definition for the two subsequences).

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    @JyrkiLahtonen haha, that is awesome. I went for a modified version of Hagens solution: suppose both series are in c, $b_0$ is found at the point $n$. a has to be right of all of the bs, because theres not enough room to the left. However, $a_0$ is at $\tilde{n}\gt n$, and therefore,$b$is finite, which is a contradiction2012-12-26