Open subset of $\ell^1$

Question

How would I go about showing that the following set $$A =\left\{x \in \ell^1 | x_1 < \sum_{n=1}^\infty x_n2^{-n}\right\}$$

is an open subset of $\ell^1$? (where $\ell^1$ is equipped with the norm $||x||_1 = \sum_n |x_n|$)

Answer 1

Consider $A^C$. Let, $p$ be a limit point of $A^C$ and let $\{x^{(n)}\}$ be a sequence of points in $A^C$ converging to $p$. Then, $\lim_{n\to \infty}x^{(n)}_k=p_k,\ k=1,2,\cdots$. Since $x^{(n)}\in A^C$, $$x^{(n)}_1\ge \sum_{k\ge 1}x^{(n)}_k2^{-k}\implies \lim_{n\to \infty}x^{(n)}_1\ge \sum_{k\ge 1}\lim_{n\to \infty}x^{(n)}_k 2^{-k}\implies p_1\ge \sum_{k\ge 1}p_k2^{-k}\implies p\in A^C.$$ Hence, $A^C$ is closed and hence $A$ is open.