Math 202B: Lecture 4

In this lecture we describe the lattice of subalgebras of $\mathcal{F}(X).$ Our main tool will be a discrete version of the Stone-Weierstrass theorem. Let $\mathcal{A}$ be a subalgebra of $\mathcal{F}(X)$ . We say that $\mathcal{A}$ separates $X$ if, for any distinct points $x\neq y$ in $X$ , there is a function $A \in \mathcal{A}$ such that $A(x) \neq A(y)$ . It turns out that there is only one separating subalgebra in $\mathcal{F}(X).$

Theorem 4.1. A subalgebra $\mathcal{A}$ of $\mathcal{F}(X)$ separates $X$ if and only if $\mathcal{A}=\mathcal{F}(X)$ .

Proof: One direction is clear: if $\mathcal{A}=\mathcal{F}(X)$ , then it contains all the elementary functions. If $x,y \in X$ are distinct, then $E_x(x)=1\neq 0=E_x(y).$

Now suppose we are given a separating subalgebra $\mathcal{A}$ of $\mathcal{F}(X)$ . Let $x \in X$ be an arbitrary point. By hypothesis, for each $y \in X\backslash\{x\}$ there is a function $A_{xy} \in \mathcal{A}$ such that $A_{xy}(x) \neq A_{xy}(y)$ , and by centering and scaling we can assume without loss in generality that

$A_{xy}(x)=1 \quad\text{and}\quad A_{xy}(y)=0.$

Now observe that

$E_x = \prod\limits_{y \in X\backslash\{x\}} A_{xy}.$

Since each of the factors $A_{xy}$ in this product belongs to $\mathcal{A}$ and $\mathcal{A}$ is closed under products, we obtain $E_x \in \mathcal{A}$ . Since $x$ was arbitrary, $\mathcal{A}$ contains all the elementary functions, and since $\mathcal{A}$ is closed under linear combinations it is equal to $\mathcal{F}(X)$ .

-QED

We are now ready to classify the subalgebras of $\mathcal{F}(X)$ . This is done by explicitly describing a family of subalgebras, and then using Theorem 2.3 to prove that there are no others apart from these.

Definition 4.1. A partition of $X$ is a set $\mathbf{P}$ of disjoint nonempty subsets of $X$ whose union is $X.$ Each $P \in \mathbf{P}$ is called a block of $\mathbf{P}.$

The set $\Pi(X)$ of all partitions of $X$ has a natural partial order called refinement order: we declare $\mathbf{P} \leq \mathbf{Q}$ if every block of $\mathbf{P}$ is a union of blocks of $\mathbf{Q}$ . When $\mathbf{P} \leq \mathbf{Q}$ we say that $\mathbf{P}$ is coarser than $\mathbf{Q}$ , or equivalently that $\mathbf{Q}$ is finer than $\mathbf{P}.$ In fact, $\Pi(X)$ is a lattice, and you way wish to think about its greatest lower bound and least upper bound operations, though we do not need these for our purposes. Below is the Hasse diagram of $\Pi(X)$ for $X$ a set of four points.

Now observe that associated to every partition $\mathbf{P}$ of $X$ is a subalgebra $\mathcal{A}(\mathbf{P})$ of $\mathcal{F}(X)$ , namely the set of functions on $X$ which are constant on the blocks of $\mathbf{P}.$

Problem 4.1. Prove that $\mathcal{A}(\mathbf{P})$ is a subalgebra of $\mathcal{F}(X)$ . Furthermore, show that $\mathcal{A}(\mathbf{P})$ is isomorphic to $\mathcal{F}(\mathbf{P})$ .

We are now ready for the classification of subalgebras of the function algebra $\mathcal{F}(X)$ .

Theorem 4.2. If $\mathcal{A}$ is a subalgebra of $\mathcal{F}(X),$ then $\mathcal{A}=\mathcal{A}(\mathbf{P})$ for some partition $\mathbf{P}$ of $X$ . In particular, every subalgebra of a function algebra is isomorphic to a function algebra.

Proof: The proof is by induction on the cardinality of $X$ . In the base case, $X=\{x\}$ is a singleton set, and $\mathcal{F}(X)=\mathcal{F}(\{x\})$ is a one-dimensional algebra. The unique partition of $X$ is $\mathbf{P}=\{\{x\}\},$ and $\mathcal{A}(\{\{x\}\})$ is the set of functions in $\mathcal{F}(\{x\})$ which are constant on $\{x\}$ , which is all of $\mathcal{F}(\{x\}).$

For the induction step, let $\mathcal{A} \neq \mathcal{F}(X)$ be a proper subalgebra of $\mathcal{F}(X)$ . Then, by the Stone-Weierstrass theorem, there are distinct points $x,y \in X$ such that $A(x)=A(y)$ for all $A \in \mathcal{A}$ . This means that every function in $A$ is constant on the the two-element set $\{x,y\}$ . Thus, $\mathcal{A}$ is a subalgebra of $\mathcal{A}(\mathbf{P}_{xy}),$ where $\mathbf{P}_{xy}$ is the partition of $X$ with one block $\{x,y\}$ of size two and the remaining blocks of size one. By Problem 3.2, the algebra $\mathcal{A}(\mathbf{P}_{xy})$ is isomorphic $\mathcal{F}(\mathbf{P}_{xy})$ , and $|\mathbf{P}_{xy}|=|X|-1.$ Thus, by the induction hypothesis, $\mathcal{A}=\mathcal{A}(\mathbf{Q})$ for $\mathbf{Q}$ a partition of $\mathbf{P}_{xy}.$