Continuity Between Metric Spaces

Continuity as Preservation of Nearby Points

In earlier tutorials, convergence and open sets were developed for metric spaces, and separability was connected to countable bases. Continuity brings these ideas together: a map is continuous when points sufficiently close to a given input have images close to the corresponding output. In metric spaces, this idea can be expressed using open sets or using convergent sequences. Both formulations will be useful, and each reveals a different aspect of continuity.

Let \((X,d_X)\) and \((Y,d_Y)\) be metric spaces, and let \(f:X\to Y\). An open set in \(Y\) can serve as a region of acceptable outputs. Continuity at a point says that there is an open region around the input whose points all map into that chosen output region.

This definition is phrased in terms of open sets, but balls make its meaning concrete. Since \(V\) is open and contains \(f(x)\), it contains some ball centered at \(f(x)\). The condition asks for an open region around \(x\) whose image stays within that ball. The next tutorial will give a direct metric formulation of this requirement using prescribed radii.

There is also a global version. For a subset \(V\subseteq Y\), the inverse image is \(f^{-1}(V)=\{x\in X:f(x)\in V\}\). Continuity can be characterized by what happens to inverse images of open sets.

Proof. Suppose first that \(f\) is continuous at every point, and let \(V\subseteq Y\) be open. If \(x\in f^{-1}(V)\), then \(f(x)\in V\). Continuity at \(x\) gives an open set \(U_x\subseteq X\) with \(x\in U_x\) and \(f(U_x)\subseteq V\). Consequently, \(U_x\subseteq f^{-1}(V)\). Every point of \(f^{-1}(V)\) therefore has an open set around it contained in \(f^{-1}(V)\), so \(f^{-1}(V)\) is open. This also holds if \(f^{-1}(V)\) is empty, since the empty set is open.

Conversely, suppose the inverse image of every open subset of \(Y\) is open in \(X\). Fix \(x\in X\), and let \(V\subseteq Y\) be open with \(f(x)\in V\). The set \(f^{-1}(V)\) is open by assumption and contains \(x\). Taking \(U=f^{-1}(V)\), we have \(x\in U\) and \(f(U)\subseteq V\). This is continuity at \(x\). Since \(x\) was arbitrary, \(f\) is continuous everywhere. \(\square\)

The theorem explains why the two definitions express the same idea. Pointwise continuity controls the image of a suitably chosen region around each input. The open-set formulation collects those local statements into one condition about inverse images.

Continuity and Sequences

A second characterization uses only convergent sequences. It is particularly convenient when a map has a simple effect on distances or when a candidate discontinuity can be tested by approaching a point along a sequence.

Proof. Suppose \(f\) is continuous at \(x\), and let \(x_n\to x\). To show that \(f(x_n)\to f(x)\), take any open set \(V\subseteq Y\) containing \(f(x)\). By continuity at \(x\), there is an open set \(U\subseteq X\) with \(x\in U\) and \(f(U)\subseteq V\). Since \(U\) is open and contains \(x\), it contains a ball \(B_r(x)\) for some \(r>0\). Convergence \(x_n\to x\) implies that \(x_n\in B_r(x)\subseteq U\) for all sufficiently large \(n\). Thus \(f(x_n)\in V\) for all sufficiently large \(n\). In particular, taking \(V\) to be any ball centered at \(f(x)\) proves \(f(x_n)\to f(x)\).

For the converse, suppose \(f\) is not continuous at \(x\). Then there is an open set \(V\subseteq Y\) containing \(f(x)\) such that no open set \(U\subseteq X\) containing \(x\) satisfies \(f(U)\subseteq V\). For each positive integer \(n\), the ball \(B_{1/n}(x)\) is an open set containing \(x\). Hence it contains some point \(x_n\) for which \(f(x_n)\notin V\). We have

so \(x_n\to x\). But \(f(x_n)\notin V\) for every \(n\). Since \(V\) is open and contains \(f(x)\), convergence to \(f(x)\) would require \(f(x_n)\) to belong to \(V\) eventually: \(V\) contains a ball around \(f(x)\), and convergence eventually places the sequence in that ball. This contradiction shows that \(f\) must be continuous at \(x\). \(\square\)

The proof also gives a useful way to demonstrate failure of continuity. It is enough to find one sequence approaching \(x\) whose images do not approach \(f(x)\). The sequence need not be the only way to approach \(x\); one failing sequence already rules out continuity.

Worked Examples

Composition and the Role of the Metric

Continuity is stable under composition. This is essential when a complicated map is built from simpler maps: if each stage is continuous, the entire construction remains continuous.

Proof. Let \(W\subseteq Z\) be open. Since \(g\) is continuous, \(g^{-1}(W)\) is open in \(Y\). Since \(f\) is continuous, the inverse image under \(f\) of this open set is open in \(X\). Inverse images under compositions satisfy

Therefore \((g\circ f)^{-1}(W)\) is open in \(X\). The Pointwise and Open-Set Continuity Theorem implies that \(g\circ f\) is continuous. \(\square\)

A common pitfall is to treat continuity as a property of a formula alone. It depends on the metrics, because the metrics determine which sets are open and which sequences converge. The earlier theorem “A Bounded Distance with the Same Convergence” gives one useful illustration. If \(d\) is a metric and \(\rho(x,y)=d(x,y)/(1+d(x,y))\), then \(d\) and \(\rho\) have the same convergent sequences. The identity map in either direction between these metric spaces is continuous by the Sequential Characterization of Continuity. This conclusion uses the established convergence result; it does not say that arbitrary changes of metric preserve continuity.

The open-set and sequential perspectives complement one another. Inverse images are often efficient for proving global continuity and for handling compositions. Sequences are often efficient for checking continuity at a point or constructing a counterexample. In metric spaces, the two methods agree exactly, as the two characterization theorems show.