Closed set ($X\setminus F\in \tau$)

A subset $F$ of a topological space $(X,\tau )$ is closed when its complement in that same ambient space is open: $X\setminus F\in \tau$. For example, the closed interval $[0,1]$ is closed in the usual real line because its complement is the open set $(-\infty ,0)\cup (1,\infty )$. Always specify the ambient topological space: a subset can be closed in a subspace topology without being closed in the surrounding space. ^[ref]

Being closed is not the logical negation of being open. In the usual real line, all four combinations occur: $(0,1)$ is open but not closed; $[0,1]$ is closed but not open; the half-open interval $[0,1)$ is neither open nor closed; and both the empty set and the whole real line are open and closed, hence clopen. Nor does being closed imply being a bounded set: $[0,\infty )$ is closed because its complement $(-\infty ,0)$ is open, but it is not a bounded set. The symbol $\infty$ is not an extra point that this interval must contain.

The coloured strokes mark membership; filled endpoint circles mean included points, and hollow endpoint circles mean excluded points. The empty set has no coloured points, while arrows at both ends of the coloured real line indicate that membership continues indefinitely. These examples use the usual topology of the real line.

In a metric space $X$, a subset $F$ is closed exactly when every convergent sequence $(x_n)$ in $F$ whose limit of a sequence $x$ lies in $X$ has $x\in F$. For example, $x_n=1-1/n$ lies in $[0,1)$ for $n\ge 1$ and has limit of a sequence $1$, which is missing; thus this half-open interval is not closed in the real line. This criterion is useful for keeping limiting constructions inside a prescribed set. ^[ref]

In every topological space, the general criterion is $F=\overline{F}$, using closure; equivalently, $F$ contains all its limit points. A closed set is always a sequentially closed set, but the converse can fail outside metric spaces. Arbitrary intersections and finite unions of closed sets are closed, by taking complements in the topology axioms.

The property of being a closed subset in a specified ambient topological space. In a metric space, a subset has this property exactly when it contains the limits of all its convergent sequences whose limits lie in the ambient space.

A sequentially closed set is one that contains every limit of its convergent sequences. This is weaker than being a closed set in a general topological space, but the two conditions agree in metric spaces.

A limit point of a subset $A$ of a topological space is a point $x$ whose every neighbourhood meets $A\setminus \{x\}$. Thus $1$ is a limit point of $[0,1)$ in the real line, even though $1\notin [0,1)$. In contrast, an isolated point of $A$ belongs to its closure but is not a limit point of $A$. This distinction separates limit points from the more inclusive notion of an adherent point.

A subset of the plane can have three successive strict changes under taking its set of accumulation points. For example letPut $B=\{(1/n+1/(n^{2}m),0):n,m\ge 1\}$ and $C=\{(1/n,0):n\ge 1\}$. Then $A^{\prime }=B\cup C\cup \{0\}$, $A^{\prime \prime }=C\cup \{0\}$ and $A^{\prime \prime \prime }=\{0\}$, where $0$ denotes the origin. For fixed $n,m$, increasing $k$ gives a point of $B$; increasing $m$ gives a point of $C$; increasing $n$ gives the origin. These are all possible limits: in any convergent sequence of distinct points, either $n$ is unbounded, or a subsequence fixes $n$ and has unbounded $m$, or a subsequence fixes both and has unbounded $k$. The first-coordinate cluster intervals $[1/n,1/n+1/n^2]$ are disjoint and separated away from the origin. Repeating this classification after each differentiation proves the displayed derived sets. Each derivative removes one layer of isolated points, and all three displayed changes are strict.