We provide $\{0,1\}$ with the discrete topology and view the product space $X:=\{0,1\}^{\mathbb{N}}:=\prod_{i\in\mathbb{N}}\{0,1\}$ (the space of every 0-1-sequence), with $0\notin\mathbb{N}$ .
Show that a base of the product topology is constituted by every set of the following form:
$U_{y_1,..., y_n}:=\{x\in X: x_1=y_1,\dotso, x_n=y_n\}\quad (n\in\mathbb{N}, y_1,\dotso, y_n\in\{0,1\})$
Hello,
I want to solve this task. To show that $U_{y_1,..., y_n}$ is a base, I have to proof, that every set in the product topology is union of elements in $U_{y_1,..., y_n}$.
I have a question about the union of these sets. For example: When I have the sets $U_{0,0,\dotso, 0}$ and $U_{1,1,\dotso, 1}$, what is $U_{0,0,\dotso, 0}\cup U_{1,1,\dotso, 1}\stackrel{?}{=}U_{1,1,\dotso, 1}$
Actually this task seems trivial to me, since every set $U_{y_1,..., y_n}$ describes a 0-1-sequence. So for every 0-1-sequence we already have a set $U_{y_1,..., y_n}$ which describes it.
Am I wrong? What do I have to show? I would be thankful for a hint.
Thanks in advance.
First, let's consider $U_{0,0,...,0}\cup U_{1,1,...,1}$. The set $U_{0}$ consists of all sequences that begin with a 0. Similarly, the set $U_1$ consists of all sequences that begin with a 1. Their union $U_0\cup U_1$ is therefore all sequences that begin with a 0 or 1, i.e., the whole space, but this is not a general phenomenon. Now consider $U_{0,0}\cup U_{1,1}$, which is the set of all sequences beginning with two 0's or two 1's. So, for example, the sequence $(0,1,0, ...)\not\in U_{0,0}\cup U_{1,1}$, but $(0,0,0,...)\in U_{0,0}\cup U_{1,1}$. Notice also that $(0,0,0,...)\not\in U_{1,1}$, so $U_{0,0}\cup U_{1,1}\not=U_{1,1}$.
Indeed, we need to show that any open set in the product topology can be written as a union of the sets $U_{y_1, ..., y_n}$, which I will call cylinder sets. As you know, open sets in the product topology here are unions of $\prod_{i \in \mathbb{N}} V_i$ where each $V_i\subset \{0,1\}$ and where the set $I=\{i\in \mathbb{N} : V_i\not=\{0,1\}\}$ is finite. Here, this means that we need to be able to, given $J\subset \mathbb{N}$ with $|J|<\infty$, and a function $f:J\rightarrow \{0,1\}$, write the collection $V=\{x \in X: x_j=f(j), \ \forall j\in J\}$ as a union of cylinder sets.
That is, we need to find finite sequences $X_1, ..., X_k$ such that $V=\cup_{m=1}^k U_{X_m}$. Here, I'm thinking of a sequence $X_m= x_{m, 1}, x_{m, 2}, ..., x_{m, n_m}$, and $$U_{X_m}=U_{x_{m, 1},\ \ x_{m, 2} \ \ , ..., x_{m, n_m}}\ \ $$
So, using the notation from above, if $J=\{1,2, ..., n\}$, then $V=U_{f(1), f(2), ..., f(n)}$ is trivial! However, what if $J$ is more complicated, like if $J=\{4,5,6, 101,102, 103\}$?
I hope that this will get you started!