Heine theorem

In several contexts, the topology of a space is conveniently specified in terms of limit points. This is often accomplished by specifying when a point is the limit of a sequence. Still, for some spaces that are too large in some sense, one specifies also when a point is the limit of more general sets of points indexed by a directed set, known as nets. A function is (Heine-)continuous only if it takes limits of sequences to limits of sequences. In the former case, preservation of limits is also sufficient; in the latter, a function may preserve all limits of sequences yet still fail to be continuous, and preservation of nets is a necessary and sufficient condition. In detail, a function f:X\to Y is sequentially continuous if whenever a sequence \left\{x_n\right\} in X converges to a limit x, the sequence \left\{f\left(x_n\right)\right\} converges to f\left(x\right). Thus, sequentially continuous functions "preserve sequential limits." Every continuous function is sequentially continuous. If X is a first-countable space and countable choice holds, then the converse also holds: any function preserving sequential limits is continuous. In particular, if X is a metric space, sequential continuity and continuity are equivalent. For non-first-countable spaces, sequential continuity might be strictly weaker than continuity. (The spaces for which the two properties are equivalent are called sequential spaces.) This motivates the consideration of nets instead of sequences in general topological spaces. Continuous functions preserve the limits of nets, and this property characterizes continuous functions. ==Formal statement==

Consider the case of real-valued functions of one real variable: {{Math proof|title=Proof|proof=(Sufficiency) Assume that f:A\subseteq\mathbb{R}\to\mathbb{R} is continuous at x_0 (in the sense of \varepsilon-\delta continuity). Let \left\{x_n\right\} be a sequence converging at x_0 (such a sequence always exists, for example, x_n=x_0 for all n\in\mathbb{N}). Since f is continuous at x_0, then \forall\varepsilon>0\exists\delta>0\forall x\left(0 For any \delta that satisfies conditions above, one can always find a natural number \nu such that for arbitrary n>\nu, \left|x_n-x_0\right| holds, since \left\{x_n\right\} converges at x_0. Combining this with expression (*), then \forall\varepsilon>0\exists\nu\in\mathbb{N}\forall n\in\mathbb{N}\left(n>\nu\implies\left|f\left(x_n\right)-f\left(x_0\right)\right| According to the definition of limit of a sequence, the expression above equivalents to \left\{f\left(x_n\right)\right\}, id est f is sequentially continuous at x_0. (Necessity) Assume that f is sequentially continuous, and f is not continuous at x_0, namely \exists\varepsilon>0\forall\delta>0\exists x\left(0 Then one can take \delta=\frac{1}{n}, where n is an arbitrary natural number, and call the corresponding point x_n:=x. In this way we have defined a sequence \left\{x_n\right\} such that \exists\varepsilon>0\forall n\in\mathbb{N}\left(\left|x_n-x_0\right| By the construction above, f\left(x_n\right)\to f\left(x_0\right) is false when x_n\to x_0, which contradicts the hypothesis of sequential continuity. Hence the proposition "If f is sequentially continuous at x_0, then f is continuous at x_0" holds. \blacksquare}} == References ==