A student asked me if $\pi$ contains itself. This can be interpreted to mean that $\pi-10^{-k}\pi=\bar\pi(k)$ for some positive integer $k$, where $\bar\pi$ is the sequence $(3,3.1,3.14,3.141,3.1415,\dots)$, which is false as it would imply that $\pi$ is rational. It can also be interpreted to mean that the sequence $\hat\pi=(3,1,4,1,5,9,2,6,\dots)$ has a proper sub-sequence that is equal to it. This will be my interpretation in this post.
Lemma. Every sequence $s:\mathbb N\to S$ with $|s(\mathbb N)|<\infty$ has a proper sub-sequence that is equal to it.
Proof. Let $F=\{y\in S:|s^{-1}(\{y\})|<\infty\}$, let $m=\max(\bigcup_{y\in F}s^{-1}(\{y\}))$, and let $n(i)=i$ for every integer $i$ with $0\leq i\leq m$, where $\max(\varnothing)=0$. Since $|S\setminus F|>0$, it is the case that an $n(i)$ can always be found such that $s(n(i))=s(i)$ and $n(i)>n(i-1)$ for every integer $i$ with $i>m$. Then $s=s\circ n|_\mathbb N$. $\square$
Alternative Proof. Let $n(1)$ be such that $n(1)\geq1$ and $s(n(1))=s(1)$. Let $n(2)$ be such that $n(2)>n(1)$ and $s(n(2))=s(2)$. Let $n(3)$ be such that $n(3)>n(2)$ and $s(n(3))=s(3)$. Repeat this process indefinitely. If this is not possible, then there is an $i$ such that $s(i)\neq s(j)$ for every $j$ with $j>n(i-1)$, i.e., $s(i)$ does not repeat indefinitely. In this case, let $n(k)=k$ for every $k\in\{1,2,\dots,i\}$ and repeat this process starting at $i+1$. Since at least one element in the range of $s$ repeats indefinitely, it is possible to repeat this process indefinitely. It follows that $s=s\circ n$. $\square$
Claim. $\hat\pi$ has a proper sub-sequence that is equal to it.
Proof. Since $\hat\pi:\mathbb N\to\{0,1,\dots,9\}$ is an infinite sequence with finite range, it follows from the above lemma that $\hat\pi$ has a proper sub-sequence that is equal to it. $\square$
Although this result is not profound, it led me to ask myself: is the fiber of every $y\in\{0,1,\dots,9\}$ under $\hat\pi$ infinite? This may be an open problem.
Happy $\pi$ Day!