Show that $\{ x = (x_n) \in H^\infty: |x_k| < 1, k = 1, …, N\}$ is open in $H^\infty$

Question

Let $H^\infty$ be the hilbert cube, with metric $d(x,y)=\sum_{n=1}^{\infty}\frac{x_n-y_n}{2^n}$

Show that $\{ x = (x_n) \in H^\infty: |x_k| < 1, k = 1, …, N\}$ is open in $H^\infty$

I think you have to use the characterisation of an open set in terms of sequences, but for the rest I have no clue of how to proof this.

Answer 1

Define $f:H^{\infty} \to \mathbb R^{N}$ by $f(x_n)=(x_1,x_2,...,x_N)$. Then $f$ is continuous and the given set is the inverse image under $f$ of $\{x\in \mathbb R^{N}: -1<x_i<1(1\leq i\leq N)\}$ so it is open.