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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?

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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

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

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    $\begingroup$ I conjecture that the OP intends $B$ to be the 4-element Boolean algebra, not the Boolean algebra of subsets of a 4-element set. Specifically, he treats 1 as the unit when he says that ultrafilters must contain it. So I think he intends 0 and 1 to be the top and bottom elements of the Boolean algebra and intends 2 and 3 to be the two atoms. $\endgroup$ – Andreas Blass Sep 8 '16 at 1:26
  • $\begingroup$ yes, you are right. Sorry for not being clear. $\endgroup$ – Hilario Fernandes Sep 8 '16 at 1:28
  • $\begingroup$ @Andreas: Yes, I should have realized that; thanks. $\endgroup$ – Brian M. Scott Sep 8 '16 at 1:38
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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.

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  • $\begingroup$ Ahh, yes, that makes much more sense. In fact, $\{ \{1,2 \} \} \cup \{\{1,3 \}\}=S(B)$. $\endgroup$ – Hilario Fernandes Sep 8 '16 at 1:37

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