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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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One of the sequences could be a suffix of the other. For example $$a_0a_1a_2\cdots=012345678901234567890123456789\cdots$$ and $$b_0b_1b_2\cdots=123456789012345678901234567890\cdots$$ is its subsequence. Both appear in $c=a$. Or did you mean to ask something else? –  Jyrki Lahtonen Dec 26 '12 at 9:52
    
@JyrkiLahtonen True, this is a special case where periodic series overlap infinitely. But I meant generally, for any 2 series a and b –  CBenni Dec 26 '12 at 9:55
    
Ok. I guessed as much, but the phrase if there can be 2 infinite... left open the possibility that we get to pick the two sequences. To give a precise answer (or a proof) we need to know you definition of a sequence. I agree with Hagen that here it looks like a sequence is any function from the set of non-negative integers to the set of digits. If we agree on that definition, then the answer is NO, because the digits must all have non-negative integers as their position index. –  Jyrki Lahtonen Dec 26 '12 at 10:02
    
Please note the difference between sequence and series: we have that $$\{x_n\}_{n\in\Bbb N}\,\,,\,x_n\in\Bbb R\,\,or\,\,\Bbb C\,\,,\text{say}$$ is a sequence, whereas $$\sum_{n=1}^K x_n$$ is a series, finite if $\,K\in\Bbb N\,$ , infinite otherwise. –  DonAntonio Dec 26 '12 at 12:10
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Don't worry, it happens to most of us here and there. –  DonAntonio Dec 26 '12 at 14:24
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up vote 2 down vote accepted

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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Yeah they should get the first part. Could have thought about that myself >_< Thank you! EDIT: well, mmh. you assume that there cannot be another infinite series after all of the $b$s. This is clear to anyone who has the slightest clue of mathematics, but apparently, these computer science people dont. –  CBenni Dec 26 '12 at 9:57
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@CBenni: The way I teach this kind of things to non-math majors is that in a sequence such as $a_0a_1\cdots$ you can single out any digit in the sequence, ask it about its position in the sequence, and it should be able to reply. Normally the algorithm is that it can tap on the shoulder of the digit in front of it, ask the same question, and add one to the answer it gets. By induction, all the digits can answer in a finite time. If you ask this question from $b_0$ in $a_0a_1\cdots b_0b_1\ldots$, it won't reply, because whose shoulder could it tap? Or even if it could the algorithm never stops. –  Jyrki Lahtonen Dec 26 '12 at 10:08
    
@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 contradiction –  CBenni Dec 26 '12 at 10:37
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