Basis of the topology of a Stone space

Question

I'm trying to understand the general idea of Stone Representation Theorem (or, at least, the existence of a functor from the category of boolean algebras to the category of compact totally disconnected spaces) with the example $B=\{0,1,2,3\}$.

The corresponding Stone space contains all ultrafilters on $B$ (which are all subsets of $B$ that contains $1$ and exactly $a$ or $¬a$, $a \in B$), so $S(B)=\{ \{1,2\},\{1,3\} \}$. Also, the basis of the space is $\beta_{S(B)}= \{ U \in S(B) \mid b \in U\} = \{ \emptyset,\{1,2\},\{1,3\} \}$. But that makes no sense, since the union of all elements of $\beta_{S(B)}$ is not an element of $S(B)$.

What am I missing, or misunderstanding?

Answer 1

In that case $S(B)$ is simply the discrete two-point space. The points are $\{1,2\}$ and $\{1,3\}$, and the open sets are $\varnothing,\big\{\{1,2\}\big\},\big\{\{1,3\}\big\}$, and $S(B)$ itself. The base is

$$\beta_{S(B)}=\big\{\{U\color{red}{\subseteq}S(B):b\in U\}:b\in B\big\}\;.$$

Here

$$\begin{align*} \{U\subseteq S(B):0\in U\}&=\varnothing\\ \{U\subseteq S(B):1\in U\}&=S(B)\\ \{U\subseteq S(B):2\in U\}&=\big\{\{1,2\}\big\}\;,\text{ and}\\ \{U\subseteq S(B):3\in U\}&=\big\{\{1,3\}\big\}\;. \end{align*}$$

Answer 2

You seem to be confusing elements of $S(B)$ with subsets of $S(B)$. Specifically, your basis should consist of four subsets of $S(B)$, one for each $b\in B$. For $b=0,1,2,3$, the corresponding subsets are $\varnothing$, $S(B)$, $\{\{1,2\}\}$, and $\{\{1,3\}\}$, respectively.