Closure, interior and boundary

Try

For A=(0,1] in the usual real line, the interior is (0,1), the closure is [0,1], and the boundary consists of zero and one. At zero, every neighborhood touches A but zero is absent from A. At one, every neighborhood touches both A and its complement even though one belongs to A.

Membership and boundary status are therefore separate questions. A boundary point need not be outside the set, and a set need not be open or closed.

Predict

Predict the closure and interior of Q in R. Every real interval meets Q, so its closure is all of R. Every interval also contains an irrational, so Q has empty interior. Its boundary is therefore the whole real line.

This example warns against treating boundary as a thin curve in a picture. Topological boundaries can be as large as the whole space.

Experiment

Compute the three operations in the usual real line.

Set	Interior	Closure	Boundary
(0,1]	(0,1)	[0,1]	0 and 1
Q	Empty	R	R
R	R	R	Empty

The ambient space matters. In the subspace [0,1], the whole set [0,1] has empty boundary, although its boundary in R has two points.

Understand

The interior of A is the union of all open sets contained in A. The closure of A is the intersection of all closed sets containing A. These definitions give the largest open subset and smallest closed superset respectively.

A point x lies in the closure precisely when every open neighborhood of x meets A. Indeed, a neighborhood missing A has a closed complement containing A but excluding x. Conversely, a closed superset excluding x supplies just such a neighborhood by complementation. This argument works in every topology and requires no sequences.

The boundary is closure(A) minus interior(A), equivalently closure(A) intersected with closure(X without A). To prove the equivalence, x is not interior to A exactly when every neighborhood meets its complement. Thus a boundary point is one whose every neighborhood meets both sides.

Closure is monotone, contains A, and is idempotent because closure(A) is already closed. It preserves finite unions: closure(A)∪closure(B) is a closed superset of A∪B, while monotonicity gives the reverse containment. Infinite unions are different. The closed singletons 1/n have a union omitting zero, but its closure contains zero.

For A⊆Y⊆X, closure within Y equals Y intersected with closure within X. A relative neighborhood Y∩U meets A exactly when U does, using A⊆Y. This proves the formula directly and avoids guessing from a drawing.

Apply

In the three-point chain topology, compute the closure of the singleton containing b. The closed supersets force the pair b,c. Explain through neighborhoods why c enters the closure even though it is a different point.

Repair “A set is closed if it contains the limits of all its sequences” as a universal topological claim. It is valid in metric spaces and more generally first-countable spaces, but may fail in arbitrary spaces. We will prove the correct scope and replace sequences by nets later.

Continue

Use the neighborhood criterion whenever an algebraic closure calculation becomes opaque. The next lesson uses it to characterize continuity through f(closure A)⊆closure(f(A)). Optional depth: derive interior(A)=X without closure(X without A), carefully keeping all complements relative to the same ambient space.