My book1 says: Let K be the set of real numbers that are sums of the series of the form $\sum_{k=1}^{\infty} \frac{a_k}{3^k}$ with ${a_k} \in \{0, 2\}$.
I can see that sum where ${a_k}=0$ goes to $0$: $$\sum_{k=1}^{\infty} \frac{0}{3^k} = \frac{0}{3^1}+\frac{0}{3^2}+\frac{0}{3^3}+\frac{0}{3^4}.... = 0$$ and that the other sum for ${a_k}=2$ goes to $1$: $$\sum_{k=1}^{\infty} \frac{2}{3^k} = \frac{2}{3^1}+\frac{2}{3^2}+\frac{2}{3^3}+\frac{2}{3^4}.... = 1$$ Those are the only sums I see, since $a_k$ can only be $0$ or $2$. I am clearly reading the notation incorrectly. Can anyone help explain what the correct way to read the notation is?
1 Elementary Topology Problem Textbook by O. Ya. Viro, O. A. Ivanov, N. Yu. Netsvetaev, V.M. Kharlamov