# Hypothetical contradiction to Bolzano-Weierstrass

we've learned about the Bolzano-Weierstrass theorem that states that if a sequence is bounded, then it has a subsequence that converges to a finite limit. Let's define $a_n$ as the digits of $\pi$, i.e. $a_1$ = 3, $a_2$ = 1, $a_3$ = 4, and so on infinitely. Certainly this sequence is bounded by 10 and 0, but I can't think of any subsequence that will converge to anything. Can you help solve my confusion?

• Just because you can't explicitly give an example does not mean one doesn't exist. – David Mitra Dec 13 '14 at 13:31
• But, see this. – David Mitra Dec 13 '14 at 13:33

Sure, take the subsequence of every occurrence of $1$. If there aren't infinitely many $1$s, then $2$s; if not $2$s then $3$s, etc. As $\pi$ is irrational, at least one non-zero digit in the decimal expansion is repeated infinitely many times.