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Call a function $f : \mathcal{N} \rightarrow \mathcal{N}$ repetitive if for every finite sequence of natural numbers $(a_1, a_2, \cdots,a_n)$ there exists a number $k \in \mathcal{N}$ satisfying

$f(k) = a_1$, $f(k+1) = a_2$, ... , $f(k + n -1) = a_n$.

Show that if $f$ is repetitive, then for any combination of $a_i$'s there are actually infinitely many numbers $k$ with this property.

I've been trying to figure out a proof for this for a while now. I'm feeling that the fact that the number $k$ exists for every sequence allows us to manipulate other sequences or something like that and get infinitely many numbers that way. But I haven't been able to actually find a way to construct infinitely many numbers $k$ for a random sequence.

hmakholm left over Monica
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xzeo
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2 Answers2

27

There's a $k$ with $f(k)=a_1,\ldots,f(k+n-1)=a_n$, $f(k+n)=1$.

There's a $k'$ with $f(k')=a_1,\ldots,f(k'+n-1)=a_n$, $f(k'+n)=2$ etc.

Angina Seng
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11

For every finite sequence there are infinitely many finite distinct sequences that extend it.

Henno Brandsma
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