Open set in Hilbert Cube. Any open set in the Hilbert Cube is the union of open subsets of the form $$U_1 \times ... \times U_n \times X_{n+1} \times .... \times X_{n+k} \times...$$
where $X_k := [0, \frac{1}{k}]$ for $k \in \Bbb N$ and where $U_i$ are open in $X_i$, $1 \leq i \leq n$  and $n$ may vary.
I cannot understand the above form of an open set in Hilbert Cube. Please Help!!
 A: This is pretty much by definition. The Hilbert cube is defined to be the Cartesian product of the intervals $[0,1/n]$. 
The topology used is the product topology which is defined in that way.

In $\prod X_i$ you put the topology in which the open sets are of the form $$\prod U_i$$ where all $U_i=X_i$ except for finitely many $i$'s.

In case that what is confusing in your problem is that they wrote the factors that are equal to $X_i$ are only in the tail, notice that $n$ is any number and that you can still have some $U_i=X_i$ at the beginning. 

But suppose you started by defining the topology through a metric. Suppose that you defined the Hilbert cube as the set $\prod_n[0,1/n]$ with the metric 
$$d(x,y)=\left(\sum (x_i-y_i)^2\right)^{1/2}$$
Consider a ball $B(x,r)$ with center $x$ and radius $r>0$. Then notice that if $N$ is such that $\sum_{n>N}\frac{1}{n^2}<r$ (which it exists because $\sum \frac{1}{n^2}<\infty$) then
$$\prod_{i=0}^{N}\{x_i\}\times\prod_{n>N}[0,1/n]\subset B(x,r)$$
This shows, that starting at some index all the factors of $B(x,r)$ are the whole interval $[0,1/n]$.
