Theorem 1

Let X be a topological space, A a dense subset of X, f : A → Y a mapping of A into a regular space Y. A necessary and sufficient condition for f to extend to a continuous mapping $\overline{f} : X \to Y$ is that, for each $x \in X$, $f(y)$ tends to a limit in Y when y tends to x while remaining in A. The continuous extension $\overline{f}$ of f to X is then unique.

The uniqueness of $\overline{f}$ follows from the principle of extension of identities (no. 1, Proposition 2, Corollary 1). It is clear that the condition is necessary, for if $\overline{f}$ is continuous on X, then for each $x \in X$ we have

(§ 7, no. 5). Conversely, suppose that the condition is satisfied and define for each $x \in X$; $\overline{f}(x)$ is a well-defined element of Y, since Y is Hausdorff. We have to show that $\overline{f}$ is continuous at each point $x \in X$. Let then V' be a closed neighbourhood of $\overline{f}(x)$ in Y; then by hypothesis there is an open neighbourhood V of x in X such that $f(V \cap A) \subset V'$. Since V is a neighbourhood of each of its points, we have for each $z \in V$, and from this it follows that $\overline{f}(z) \in \overline{f}(V \cap A) \subset V'$, since V' is closed. The result now follows from the fact that the closed neighbourhoods of $f(x)$ form a fundamental system of neighbourhoods of $f(x)$ in Y.

The mapping $\overline{f}$ is said to be obtained by extending f by continuity to X.

In the statement of Theorem 1 the hypothesis that Y is regular cannot be weakened without imposing additional restrictions on X, A or f (Exercise 19).