Discrete space

A topological space is discrete when every subset is an open set.

A subset of a topological space is discrete when its subspace topology is a discrete topology: every point has an ambient neighbourhood meeting the subset only in that point. A discrete subset need not be closed; for example $\{1/n:n\ge 1\}$ in the real line accumulates at zero. The group structure makes a discrete subgroup of a Hausdorff topological group closed, an additional property used in discontinuity arguments.

A closed discrete subset $D$ of a topological space $X$ is a closed set whose subspace topology is the discrete topology. Every point of $D$ therefore has a neighbourhood meeting $D$ at that point alone. Closedness excludes all accumulation points outside $D$ as well.

In a plane domain, a closed discrete subset is exactly a set meeting every compact subset in finitely many points. This local finiteness is necessary for a nonzero holomorphic function to have the subset as its entire zero set. A merely isolated-point subset can have an accumulation point outside the subset but inside the domain, as $\{1/n:n\ge 1\}\subset \mathbb{C}$ does.

// Destination: analysis.bigb

Let $D$ be a closed discrete subset of a compact space $X$. Closedness makes $D$ compact. Discreteness makes its singletons open in its subspace topology, so the singleton cover has a finite subcover. Thus $D$ is finite. Closedness is essential: the discrete set $\{1/n:n\ge 1\}$ inside the compact interval $[0,1]$ is infinite and has the missing accumulation point zero.

Enumerate all intervals with rational endpoints, using the countability of the rational numbers and listing pairs of enumeration indices by increasing sum. If every point $b$ of $B$ is an isolated point, some such interval contains $b$ and no other point of $B$. Assign to $b$ the least index of such an interval. Distinct points cannot have the same index, so this gives an injective function into $\mathbb{N}$ and proves that $B$ is a countable set. The argument uses isolation in the subspace topology, not closedness: the discrete subset $\{1/n:n\ge 1\}$ illustrates that an accumulation point outside $B$ is allowed.