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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.

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$(x_n)$ 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 $[0,1]$.

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    $\begingroup$ For inspiration on the rest, see my answer here. $\endgroup$ Nov 2, 2021 at 17:28
  • $\begingroup$ It took me some time to understand, but I finally managed to solve this exercise by intensively consulting the answer you provided as a link. I therefore thank you for your help and will accept your answer gratefully. $\endgroup$
    – Octavius
    Nov 8, 2021 at 20:31

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