A Stone Space $T$ is homeomorphic to $SB(T)$

Question

...where $B: Stone \to Bool$ is the functor that assigns each Stone Space to the boolean algebra of clopen sets on such space (with the usual set operations) and $S: Bool \to Stone$ is the functor that assigns each Boolean Algebra to the space of its ultrafilters, in which the basic open sets are collections of all ultrafilters which contain a certain element of the boolean algebra.

Here's an example I thought of. Consider the space $T=\{1,2,3\}$ with the discrete topology. Then, $B(T)= \mathcal P \left({T}\right)$. Consequently, $SB(T) = \{ \{T, \{1\}, \{2\}, \{3\} \},\{T, \{1\}, \{2\}, \{1,2\} \},\\ \{T, \{1\}, \{1,3\}, \{3\} \},\{T, \{1\}, \{1,3\}, \{1,2\} \},\\ \{T, \{2,3\}, \{2\}, \{3\} \},\{T, \{2,3\}, \{2\}, \{1,2\} \},\\ \{T, \{2,3\}, \{1,3\}, \{3\} \},\{T, \{2,3\}, \{1,3\}, \{1,2\} \} \}$.

Well, pretty obviously the spaces $T,SB(T)$ are not homeomorphic. What am I missing here?

Answer 1

Your computation of $SB(T)$ is incorrect. There are only three ultrafilters on $B(T)$, namely the principal ones generated by $1$, $2$, and $3$. That is, \begin{align*} SB(T) =\, &\{\{\{1\}, \{1,2\}, \{1, 3\}, T\}\\ & \{\{2\}, \{1,2\}, \{2, 3\}, T\}\\ & \{\{3\}, \{1,3\}, \{2, 3\}, T\}\}.\end{align*}

The homeomorphism $T\cong SB(T)$ sends $n$ to the principal ultrafilter generated by $n$.

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The other sets you wrote down fail to be ultrafilters because they are not closed upward or closed under intersection. For example, $U = \{T,\{1\},\{2\},\{1,2\}\}$ is not closed upward (since $\{1\}\in U$, we should have $\{1,3\}\in U$) or intersection (since $\{1\}\in U$ and $\{2\}\in U$, we should have $\emptyset\in U$).