In a topological space you already have a rule for which sets are open. Openness is the whole of the structure — and yet almost the first thing every topology does is turn that single notion inside out. Take any open set, look at everything it leaves behind, and you have a closed set. That one act of complementation unlocks a small vocabulary — interior, closure, boundary, limit point — that lets you describe exactly how a set sits inside its space: which points are comfortably inside, which are clinging to the edge, and which the set is "reaching toward" without quite containing.

A set C \subseteq X is closed when its complement X \setminus C is open. That is the entire definition — closed does not mean "not open", and it does not mean "bounded". It means one thing only: the complement is open.

The axioms, turned inside out

The axioms for open sets — the empty set and the whole space are open, arbitrary unions of open sets are open, finite intersections are open — pass through the complement and land as a mirror set of rules for closed sets. De Morgan's laws do all the work: complementing a union gives an intersection, and complementing an intersection gives a union, so every "union" above becomes an "intersection" below, and vice versa.

The word finite in the last line is not decoration. An infinite union of closed sets can fail to be closed: each singleton \{1/n\} is closed in \mathbb{R}, yet \bigcup_{n \ge 1} \{1/n\} = \{1, \tfrac12, \tfrac13, \dots\} is not closed — it is missing its limit point 0. Dually, an infinite intersection of open sets can fail to be open: \bigcap_{n \ge 1} (-\tfrac1n, \tfrac1n) = \{0\}.

Open, closed, both, or neither

Here is the trap "closed" lays for beginners: open and closed are not opposites, and they are not exhaustive. A set can be one, the other, both, or neither.

• Open but not closed: (0, 1) \subseteq \mathbb{R}.
• Closed but not open: [0, 1] \subseteq \mathbb{R}.
• Both open and closed (clopen): in \mathbb{R} only \varnothing and \mathbb{R} itself. (In a disconnected space there are more — clopen sets are precisely what a space can be split along, so "how many clopen sets?" is really the question "how connected is this space?".)
• Neither open nor closed: the half-open interval [a, b) \subseteq \mathbb{R}. It is not open — no ball around a stays inside it. It is not closed — its complement (-\infty, a) \cup [b, \infty) is not open, because no ball around b stays outside [a,b)… wait, look again: b \notin [a,b), yet [a,b) is "reaching toward" b, so b cannot be surrounded by points of the complement. Neither box is ticked.

The rest of this page is the machinery for pinning down exactly that "reaching toward": given any set at all — open, closed, or the messy neither — how do we name its inside, its edge, and the smallest closed set it lives in?

Interior: the largest open set inside

A point x is an interior point of A if A contains a whole open neighbourhood of x — there is room to wiggle in every direction and still stay in A. The interior \operatorname{int}(A) (also written A^{\circ}) collects them all. Equivalently, and this is the definition that generalises cleanly:

Because arbitrary unions of open sets are open, \operatorname{int}(A) is itself open — and it is the largest open set contained in A: any open U \subseteq A is one of the sets in the union, so U \subseteq \operatorname{int}(A). Two immediate consequences: \operatorname{int}(A) \subseteq A always, and A is open \iff \operatorname{int}(A) = A.

On the line, \operatorname{int}\big([0, 1]\big) = (0, 1): the endpoints have no room — every neighbourhood of 0 pokes out to the left of A. And \operatorname{int}(\mathbb{Q}) = \varnothing, because every interval, however tiny, contains irrationals: no rational has any breathing room inside \mathbb{Q}.

Closure: the smallest closed set outside

The closure \operatorname{cl}(A) (also written \overline{A}) is the mirror image of the interior. Dually to the union of open subsets, take the intersection of closed supersets:

Because arbitrary intersections of closed sets are closed, \overline{A} is closed — and it is the smallest closed set containing A. Dually to the interior: A \subseteq \overline{A} always, and A is closed \iff \overline{A} = A. There is a clean "outside" description too: a point x lies in \overline{A} exactly when every open neighbourhood of x meets A — x is a point that A cannot be pried away from.

Interior and closure are exchanged by complementation, the single most useful identity in the subject:

Read it aloud: the points not in the interior of A are exactly the closure of the complement. Everything you prove about interiors gives a closure theorem for free, and vice versa.

Boundary: the thin skin between inside and outside

The boundary \partial A is what closure adds on top of the interior:

A boundary point is one that every neighbourhood straddles: each ball around it meets both A and its complement. So the three sets \operatorname{int}(A), \partial A, and \operatorname{int}(X \setminus A) (the "exterior") partition X into inside, skin, and outside. Note that \partial A = \partial(X \setminus A): a set and its complement share the very same boundary.

The figure below shows a subtle set in the plane: an open disc together with a single isolated point q sitting off to the side. Watch how the interior, boundary and closure peel apart — and pay attention to where the isolated point lands.

The interior is just the open disc — the isolated point q has no room, so it is not interior. The boundary is the rim of the disc plus q itself: q \in A but every neighbourhood of it also hits the complement, so it is a boundary point that belongs to A. And the closure fills in the rim (points A was reaching toward) while q just stays as it was. This one picture breaks two beginner reflexes at once — see the "Watch out!" box below.

Limit points: what the set is reaching toward

A point x is a limit point (or accumulation point) of A if every open neighbourhood of x contains a point of A other than x itself:

The clause "\setminus \{x\}" is the whole subtlety. It forbids x from being counted merely because it is in A — the set must have other points crowding arbitrarily close. The collection of all limit points is written A', the derived set. It gives the cleanest formula for closure:

A point of A that is not a limit point is an isolated point: it has a neighbourhood meeting A only at itself. That is precisely the point q in the figure — it is in the closure (trivially, being in A) and on the boundary, but it is not a limit point. Two more sharp examples on the line:

• A = \{1/n : n \in \mathbb{N}\} has exactly one limit point, 0 — and 0 \notin A, which is exactly why A is not closed. Here \overline{A} = A \cup \{0\}.
• A finite set has no limit points at all (A' = \varnothing), so it equals its own closure: every finite set is closed.

And a set is closed iff it contains all of its limit points (A' \subseteq A) — because then \overline{A} = A \cup A' = A. This is often the most practical closedness test you own.

Dense sets: reaching toward everything

A set A is dense in X when its closure is the whole space:

The archetype is \mathbb{Q} inside \mathbb{R}. Every open interval, no matter how short, contains a rational, so \overline{\mathbb{Q}} = \mathbb{R} — the rationals are reaching toward every real at once. (Meanwhile \operatorname{int}(\mathbb{Q}) = \varnothing: \mathbb{Q} is "everywhere and nowhere", topologically thick and thin at the same time.) A space that contains a countable dense subset — as \mathbb{R} does, via \mathbb{Q} — is called separable. Separability is what lets analysts approximate every point by a countable list, and it is why \mathbb{R} is far more tractable than an arbitrary space.

The closure operator: Kuratowski's four axioms

There is a beautiful twist. We built closure from open sets. But you can run the logic backwards: hand someone only the operation A \mapsto \overline{A}, obeying four rules, and they can reconstruct the entire topology from it. These are the Kuratowski closure axioms. For all A, B \subseteq X:

Declare a set closed exactly when \overline{A} = A, call the complements open, and you recover a genuine topology whose closure operator is the one you started from. Openness and closedness are not more fundamental than closure — the three are interchangeable foundations for the very same idea.

Worked examples: computing int, cl and ∂

Example 1 — the half-open interval (0, 1] in \mathbb{R}. Strip to the largest open subset: \operatorname{int}(0, 1] = (0, 1) (the endpoint 1 has no room to the right). Add the limit points: A' = [0, 1], so \operatorname{cl}(0, 1] = (0, 1] \cup [0, 1] = [0, 1]. Hence \partial (0, 1] = [0, 1] \setminus (0, 1) = \{0, 1\}. Notice that 1 \in A is a boundary point, while 0 \notin A is also a boundary point: the boundary happily includes points of the set and points outside it.

Example 2 — the rationals \mathbb{Q} in \mathbb{R}. Every interval contains both rationals and irrationals, so no point has room inside \mathbb{Q} and none has room outside: \operatorname{int}(\mathbb{Q}) = \varnothing and \overline{\mathbb{Q}} = \mathbb{R}. Therefore \partial \mathbb{Q} = \mathbb{R} \setminus \varnothing = \mathbb{R} — the entire line is boundary. A set can be all edge and no inside.

Example 3 — the open disc D = \{(x,y) : x^2 + y^2 < 1\} in the plane. It is already open, so \operatorname{int}(D) = D. Its limit points fill in the rim: \overline{D} = \{x^2 + y^2 \le 1\}, the closed disc. The boundary is the unit circle \partial D = \{x^2 + y^2 = 1\}, a genuine one-dimensional curve bounding a two-dimensional region — the everyday meaning of "boundary" recovered as a special case of the topological one.