Infinite product space (TOPOLOGY)

Question

I am asking about the infinite product of $\{0,1\}$. That is, $\{0,1\}^\Bbb N$ is the space of all infinite sequence of $0_s$ and $1_s$, how collection of all finite sequences of $0_s$ and $1_s$ become a basis for its topology.

Why this is the case?

Answer 1

OK, let $p_n$ the projection onto the $n$-th coordinate. As you mention in the comments, a subbase for the topology on $\{0,1\}^\mathbb{N}$ is given by all sets $p_n^{-1}[O]$, where $O \subset \{0,1\}$ is open. Taking $O$ to be empty is pointless and taking $O = \{0,1\}$ is also pointless (as then the inverse image is just the whole space). So we only have to consider $O=\{0\}$ and $O = \{1\}$. So define $O(0, n) = p_n^{-1}[\{0\}]$ and $O(1,n) = p_n^{-1}[\{1\}]$.

So $O(0,n)$ is the set of all infinite sequences of zeroes and ones that have a $0$ on the $n$-th coordinate, and similarly for $O(1,n)$ and $1$.

Now, if we have a finite sequence $a_0, a_1,\ldots,a_n$ is a finite sequence of zeroes and ones, then define $O(a_0,a_1,\ldots,a_n) = \cap_i O(a_i, i)$, so the set of sequences that agree with the finite sequence on the first coordinates. All these sets are in the base generated by the subbase, as finite intersections. And if we have any finite intersection $O(a_i, i)$ for some $i \in F$, $F$ a finite subset of coordinates, which contains some $(x_n)$, then $(x_n) \in O(x_0,x_1,\ldots,x_{\max(F)}) \subset \cap_{i \in F} O(a_i, i)$, so the sets $O(a_0,\ldots,a_n)$ (for all finite sequences) is a base for the product topology as well.

This is what is meant when we say the finite sequences form a base for the product: for every finite sequence $(a_0,\ldots,a_n)$, the sets $O(a_0,\ldots,a_n)$ (all infinite sequences that agree with the finite sequence in the beginning), form a convenient base for the product topology.