# Closed subalgebras of $C[0,1]$ and $C[0,1)$?

Let $C[0,1]$ be the algebra of continuous functions $f: [0,1] \to \mathbb R$.

Let $C[0,1)$ be the algebra of bounded continuous functions $f: [0,1) \to \mathbb R$.

Can anyone give some interesting nontrivial examples of unital uniform-closed subalgebras of either? Extra points if those subalgebras are maximal subalgebras of $C(X)$!

I believe every such subalgebra $A$ is necessarily a lattice.

That means it includes the functions $\sup\{f,g\}$ and $\inf\{f,g\}$ and $| \ f|$ for every $f,g \in A$.

I also think for $X=[0,1]$ the algebra cannot separate points.

Any example also must have the following property:

Property: Suppose $g \in C(X)$ and, for every $x,y \in X$, we can find an element $f^x_y \in A \subset C(X)$ such that $f^x_y(x) = g(x)$ and $f^x_y(y) = g(y)$. Then $g \in A$ as well.

But I'm having trouble coming up with nontrivial examples. The first thing that springs to mind is the subalgebra of polynomials. But those functions are not bounded and, even considering $X=[0,1]$ the Stone-Weierstrauss theorem says the polynomials are dense, so the subalgebra is necessarily not closed! My next idea was some subalgebra of the polynomials but even then the Muntz-Sats Theorem says every subalgebra is dense too!