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A map $f:X\to Y$ between topological spaces is continuous when every open set $V\subseteq Y$ has an inverse image $f^{-1}(V)=\{x\in X:f(x)\in V\}$ that is an open set in $X$. The notation denotes a subset, so $f$ need not have an inverse function.

For the continuous map $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^2$, with the usual metric topology,The two open intervals form an open set; an inverse image can split into separate pieces without violating continuity. The defining condition concerns inverse images, rather than images: here $f((-1,1))=[0,1)$ is not an open set in the real line. Checking this one inverse image illustrates the definition; it alone does not prove continuity for all target open sets.

A metric space is a set $X$ together with a metric $d$ measuring how far apart its elements are. The elements can be positions, functions, or other objects: what matters is that the metric obeys the same basic rules. This lets one discuss convergence in a metric space, continuous maps, and Cauchy sequences without first assigning Cartesian coordinates. The useful abstraction is that many arguments require only comparisons of distances, rather than the particular objects being compared.

Even the same set can support several informative metrics. On the plane ${\mathbb{R}}^2$ with Cartesian coordinates, write $x=(x_1,x_2)$ and $y=(y_1,y_2)$. Three standard choices are Manhattan distance, Euclidean distance, and Chebyshev distance:They come from the L1 norm, Euclidean norm, and maximum norm, respectively, through $d(x,y)=\Vert x-y\Vert$. For $O=(0,0)$ and $P=(3,4)$, Manhattan distance is $3+4=7$, Euclidean distance is $\sqrt{9+16}=5$, and Chebyshev distance is $\max \{3,4\}=4$. All three answers are correct for their respective metrics.

The choice encodes the question being asked. Manhattan distance models travel along a rectangular street grid. Euclidean distance measures unrestricted travel along a line segment. Chebyshev distance measures the time needed when two Cartesian coordinates can change simultaneously, each at speed at most one. Thus choosing a metric is part of modelling, rather than a consequence of knowing the endpoints. These examples have the same metric topology, but different metric path lengths and different collections of metric geodesics.

A historical reference for the general study of metric spaces is Maurice Fréchet's 1906 paper Sur quelques points du calcul fonctionnel, published in volume 22 of the Rendiconti del Circolo Matematico di Palermo. ^[ref]

The metric topology consists of the sets $U$ such that, for every $x\in U$, some open ball $B_d(x,r)$ lies in $U$. Thus open sets describe which sufficiently small changes remain inside a region; they do not specify the numerical size of those changes.

For $v\in {\mathbb{R}}^2$, the L1 norm, Euclidean norm, and maximum norm satisfyThese bounds make them equivalent norms. For example, $B_{\infty }(a,r/2)\subseteq B_1(a,r)\subseteq B_{\infty }(a,r)$: every open ball contains a sufficiently small open ball for either other metric. Hence all three define the same open sets, convergent sequences, and continuous maps. More generally, any two norms on a real or complex finite-dimensional vector space are equivalent norms; see H. A. Priestley, Functional Analysis I, §§0.5–0.7 and Theorem 4.3.

The metric topology therefore does not determine numerical distances. Multiplying a metric by $100$ preserves its open sets while multiplying every distance by $100$. Nor does the general equivalence of metrics require constant-factor bounds: that stronger comparison holds for the three norms used here.

The norm topology is the metric topology induced by $d(x,y)=\Vert x-y\Vert$.

On two copies of the real line with their usual topologies, let $U=(1,2)$ and $V=(2,4)$. The coordinate projection maps giveThe first two open sets are unbounded strips; their set intersection is an open rectangle. The point $(1.4,3)$ passes both coordinate tests, whereas $(2,3)$ fails the first because the endpoint $2$ is excluded.

Every point of an open set in the Euclidean plane has such an open rectangle around it inside that open set. Conversely, every open rectangle contains a small Euclidean ball around each of its points. Thus the product topology here is the usual metric topology of the Euclidean plane. With finitely many factors all coordinates may be restricted; on an infinite Cartesian product a basic cylinder set must leave all but finitely many factors free.

The weighted binary-sequence metric is finite because $0\le d(ϵ,\eta )\le {\sum }_{n\ge 1}{2}^{-n}=1$. It is symmetric and nonnegative. If $d(ϵ,\eta )=0$, every nonnegative summand is zero, so every coordinate agrees; conversely agreement makes the sum zero. For three sequences, apply $|{ϵ}_n-{\zeta }_n|\le |{ϵ}_n-{\eta }_n|+|{\eta }_n-{\zeta }_n|$ in each coordinate and sum. This proves the triangle inequality, so $d$ is a metric.

If two sequences differ in one of their first $n$ coordinates, their distance is at least $2^{-n}$. Thus the open ball $B_d(\eta ,2^{-n})$ is contained in $C_n(\eta )$. For every $\eta \in C_n(ϵ)$, the latter cylinder equals $C_n(\eta )$, so each cylinder set is open in the metric topology. Every set in the specified topology, being a union of cylinder sets, is therefore metric-open.

Conversely, agreement in the first $n$ coordinates givesGiven a metric-open set containing $ϵ$, choose a ball $B_d(ϵ,r)$ inside it, then choose $n$ with $2^{-n}<r$. The bound gives $C_n(ϵ)\subseteq B_d(ϵ,r)$, establishing the required prefix-neighbourhood condition. The two topologies coincide. The strict inequality $2^{-n}<r$ avoids incorrectly treating the closed tail bound as a strict ball bound.

In a relatively compact coordinate ball the Riemannian metric norm is bounded above and below by positive multiples of the Euclidean norm. A coordinate segment gives the upper distance bound. A path leaving the ball has a positive lower length bound before its first exit; paths staying in it have the Euclidean displacement lower bound. Consequently distinct points have positive Riemannian distance, and the resulting metric topology equals the manifold topology. Connectedness makes the distance finite because smooth manifolds are locally path connected.

On $\{0,1{\}}^{\mathbb{N}}$ this sum is a metric, since each summand obeys the triangle inequality and a zero sum forces coordinatewise equality. A prefix cylinder set fixing the first $n$ coordinates contains the open ball of radius $2^{-n}$ about each of its points and has diameter at most $2^{-n}$. These bounds show that the metric topology equals the prefix cylinder set topology, the usual Cantor space topology.