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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.

An open neighbourhood of a point $x$ in a topological space $(X,\mathcal{T})$ is an open set $U$ containing $x$. A general neighbourhood need only contain an open neighbourhood; it need not itself be open. A finite intersection of open neighbourhoods of the same point is again an open neighbourhood, by the topology axioms.

An open set is a member of the topology on a topological space. In a metric space, a subset $U$ is an open set precisely when every $x\in U$ has an open ball $B(x,r)$ contained in $U$ for some $r>0$. For $U=(0,1)$, one can take $r=\frac{1}{2}\min (x,1-x)$. At $x=0.2$, this gives $r=0.1$ and $B(x,r)=(0.1,0.3)\subset U$. The radius depends on the point; no uniform radius is required. ^[ref]

Openness depends on the ambient topological space. In $A=[0,1]$ with its subspace topology, $[0,1/2)=A\cap (-1,1/2)$ is an open set in $A$, although it is not an open set in the real line. Similarly, the whole closed interval $A$ is both an open set and a closed set in itself. These relative open sets are what the continuity test uses for maps involving a subset with its subspace topology.

Arbitrary intersections of open sets need not be open sets: ${\bigcap }_{n\ge 1}(-1/n,1/n)=\{0\}$, a singleton that is not an open set in the usual real line. Any point $x\ne 0$ is excluded once $n>1/|x|$. This is why the topology axiom specifies finite intersections.

The product topology on $X\times Y$ consists of arbitrary unions of rectangles $U\times V$, with $U$ open in $X$ and $V$ open in $Y$. Equivalently, these rectangles form a basis.

The empty set is the empty union, and $X\times Y$ is itself an allowed rectangle. An arbitrary union of sets of this form is still a union of allowed rectangles. Finally,which is an allowed rectangle because the factor topologies are closed under finite intersections. Distributing intersection over unions shows that the intersection of any two product-open sets is product-open; induction handles all finite intersections. These verify every topology axiom. Rectangles alone are generally not a topology, which is why their arbitrary unions are included.

A metric space metric $d:X\times X\to [0,\infty )$ satisfies positivity, $d(x,y)=0$ exactly when $x=y$, symmetry, and the triangle inequality. A subset $U\subseteq X$ is an open set when every $x\in U$ has some $\epsilon >0$ with the open ball $B_d(x,\epsilon )\subseteq U$.

Both $\varnothing$ and $X$ are open. An arbitrary union of open sets is open because a point lies in one member supplying a ball. A finite intersection is open because the minimum of the finitely many available radii supplies a ball. Thus the metric-open sets satisfy the topology axioms.

For $C[0,1]$, the Cauchy-Schwarz inequality gives $d_1(f,g)\le d_2(f,g)$, so the $d_2$ topology is at least as fine as the $d_1$ topology. To see strictness, define the continuous triangular spikesThenThus $f_n$ converges to zero in the L1 norm but not in the L2 norm. The two metrics induce different topologies.

The quotient topology on $\tilde{X}$ isSince inverse images preserve arbitrary unions and finite intersections,Also ${\pi }^{-1}(\varnothing )=\varnothing$ and ${\pi }^{-1}(\tilde{X})=X$. The displayed collection therefore satisfies all the topology axioms.