Let $A=$ $\{$ $\sum_n a_n3^{-n}$ : $a_n$ $=$ $0,2$ $\}$.
Show that $A$ as a subset of $[0,1]$ is closed:
My idea consists of taking a point $x$ in the complement of $A$. However, I'm not sure then how to proceed. Can every element of [0,1] be expressed as a sum of $b_n3^{-n}$? Where $b_n=1?$ would the complement of A be of that form? May I have hints?
Please use this definition of A rather than any alternative definition.
If $X=\{0,2\}^\mathbb{N}$ has the product topology, check that $f: X \to [0,1]: f(x)=\sum_n a_n 3^{-n}$ is continuous.
As $X$ is compact, so is its image $A$. And compact subsets of $[0,1]$ are closed.