Basis of product topology Why are typical basis elements of a countably infinite product topology of $\prod_{i=1}^\infty X_i$ of the form
$$\prod_{i=1}^n U_i\times \prod_{i=n+1}^\infty X_i?$$
where of course the $U_i$ are open in the corresponding $X_i$.
Everywhere I look, it says that this is easy to see. Sadly, I just don't see it and I hope someone can help me out.
Edit: The definition of product topology that I use says that for $X\times Y$ it has as subbasis $\{U\times Y\}\cup \{X\times V\}$ with $U,V$ opens in $X,Y$ resp.
 A: If $X = \prod_{n \in \mathbb{N}} X_n$, then a subbase for this topology is given, as for any product, by all sets of the form $[n,U]:= (\pi_n)^{-1}[U]$, where $ n \in \mathbb{N}$ and $U$ is open in $X_n$, which we could also denote (slightly less acurately, abusing notation a bit) as 
$$[n,U] = \prod_{m < n} X_m \times U \times \prod_{m > n} X_m$$
So we have a set that is only not the whole space for one coordinate $n$, where it equals $U$ instead.
Now the finite intersections of all subbasic elements always form a base for the topology that is generated  by the subbase.
Now, note that 
$$\prod_{i=1}^n U_i \times \prod_{i= n+1}^\infty X_i = \bigcap_{i=1}^n[ i,U_i] $$
so these sets are all in the base.
And if we have any finite intersection of subbasic elements, say $[i, U_i], i \in F$, where $F \subset \mathbb{N}$ is finite, then set $n = \max(F)$, 
and for $i \le n$ define $V_i = U_i$ if $i \in F$ and $V_i = X_i$ otherwise, now all $V_i$ are open in $X_i$, and 
$$\bigcap_{i \in F} [i, U_i] = \prod_{i=1}^n V_i \times \prod_{i=n+1}^\infty X_i$$
is also of the right form; we just "stuff"" with $X_i$ and finite sets have a maximum, so beyond that we always get $X_i$.
So all sets of the prescribed form are exactly all finite intersections of subbasic elements, so a base for the product topology.  
