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A topological space is totally separated when any two distinct points are separated by a clopen set. Such a space has no nontrivial connected subset. A zero-dimensional space with its usual clopen-base definition and $T_1$ is totally separated. Conversely these conditions agree for compact Hausdorff spaces by the Šura-Bura lemma and compactness, although the definitions should not be conflated for arbitrary spaces. Any closed totally separated subset of the unit square is nowhere dense, since every nonempty relatively open subset of the square contains a nontrivial connected line segment. Restricting a clopen set separating the segment's endpoints would disconnect that segment.

A uniform algebra on a compact Hausdorff space $K$ is a closed complex subalgebra of $C(K)$, with the supremum norm, containing the constants and separating points. Evaluation embeds $K$ continuously and injectively in the character space of an algebra. The embedding is a homeomorphism onto a closed subset because its domain is compact and its target is Hausdorff. Extra characters need not be point evaluations on the specified $K$.

For an algebra $A\subseteq C(K,\mathbb{F})$ on a compact Hausdorff space $K$, with $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$, its uniform closure consists of the uniform limits of sequences from $A$. It is a closed algebra in the supremum norm, since uniformly convergent sequences are uniformly bounded andFor a real algebra containing the constants and separating points, the Stone-Weierstrass theorem makes this closure all of $C(K,\mathbb{R})$. For a complex algebra, closure under complex conjugation is additionally sufficient; separation and constants alone are not enough.^[ref]

For example, complex polynomials in $z$ on the closed unit disk contain constants and separate points, but their uniform closure is the disk algebra, not all continuous functions on that disk. They cannot approximate $\overline{z}$ uniformly: uniform limits of these holomorphic functions are holomorphic in the interior, while $\overline{z}$ is not.