Open sets in the infinite product of $\mathbb{N}^\infty$

Question

How can I see that the open sets in the infinite product of $\mathbb{N}^\infty$ in the product topology with all components being the discrete topology are exactly the sets extending the given finite sequence ?

Answer 1

The standard base for the product topology on $\Bbb N^\infty$ is all sets of the form $\prod_{n \in \omega} U_n$ where all $U_n \subseteq \Bbb N$ and all but finitely many $U_n$ are equal to $\Bbb N$ (i.e. $\{n \in \omega: U_n \neq \Bbb N\}$ is a finite set).

But we can WLOG assume that the finitely many proper subsets are all singletons and occur as the initial part of $\omega$ and then we still have a base, which is smaller (call such basic open sets "simple basic sets" for now):

Let $U=\prod_n U_n$ be a basic subset as described in the beginning and let $(x_n) \in U$. Let $\{n \in \omega: U_n \neq \Bbb N\}$ have $N$ as its maximum element.

Then define $V:= \prod_n V_n$ to be the basic subset with $V_n = \{x_n\}$ for all $n \le N$ and $V_n = \Bbb N$ otherwise. This by construction is a simple basic set and $(x_n) \in V$ is obvious and so is $V \subseteq U$.

As we can do this for all $(x_n) \in U$, every standard basis set is a union of "simple basic sets" and so all open sets are unions of "simple basic sets", and these are precisely the sets of the form

$B((a_1,\ldots,a_n)) := \{(x_n) \in \Bbb N^\omega: (x_n)_n \text{ extends } a_1, a_2, \ldots a_n\}$

So you can say that $O$ in the product is open iff for each $x \in O$ there is some initial part $x|N$ of $x$, such that all sequences that extend $x|N$ are also in $O$, which is the more precise version of your "extension characterisation".