# Induced topology on Hilbert cube

I am given an exercise which I have some trouble understanding:

Let $$X$$ be the Hilbert cube, i.e. the product space $$X = [0,1]^{\mathbb{N}}$$ and let it be equipped with the product topology, thus $$B \enspace = \enspace \big\{ \; (x_n) \in H \; \big| \; \exists \, N \in \mathbb{N} \enspace \text{and open intervals} \enspace I_j \subset [0,1] \, , \; j \leq N \enspace \text{such that} \enspace x_j \in I_j \; \big\}$$ Show that the metric defined as $$d(x,y) \enspace = \enspace \sum_{n \in \mathbb{N}} 2^{-n} |x_n - y_n|$$ induces this topology, i.e. show that balls in metric $$d$$ are elements of this basis and that every element in $$B$$ is the union of balls in $$d$$.

I am a bit lost here. First of all, I am not quite sure how to understand the basis $$B$$. Does the notation $$(x_n)$$ imply a sequence of elements in $$H$$ or is it just an odd way of writing a single element of $$H$$? In the first case, does the notation $$x_j \in I_j$$ refer to the $$j$$-th component of the single element $$x \in H$$ or the $$j$$-th element of the sequence $$(x_n)$$? Why must the leading (vector/sequence)-components lie within such intervals, and what do the others do? Vanish? Why this particular definition?

Sorry, I am so confused right now, I hope you can help me out a bit. I would also be glad about some hints on how to solve this exercise.

$$(x_n)$$ is a single element of $$H$$ (you can think of it as an infinite vector). $$x_j$$ is its $$j$$-th component. There is no condition on the $$x_j$$ with $$j>N$$, they can be anything in $$[0,1]$$.