Editing Continuous function (section)

==== Sequences and nets {{anchor|Heine definition of continuity}}====
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 family|indexed]] by a [[directed set]], known as [[Net (mathematics)|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 <math>f : X \to Y</math> is '''[[Sequential continuity|sequentially continuous]]''' if whenever a sequence <math>\left(x_n\right)</math> in <math>X</math> converges to a limit <math>x,</math> the sequence <math>\left(f\left(x_n\right)\right)</math> converges to <math>f(x).</math>  Thus, sequentially continuous functions "preserve sequential limits."  Every continuous function is sequentially continuous. If <math>X</math> is a [[first-countable space]] and [[Axiom of countable choice|countable choice]] holds, then the converse also holds: any function preserving sequential limits is continuous. In particular, if <math>X</math> 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 space]]s.) 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.

For instance, consider the case of real-valued functions of one real variable:<ref>{{cite book |title=Calculus and Analysis in Euclidean Space |edition=illustrated |first1=Jerry |last1=Shurman |publisher=Springer |year=2016 |isbn=978-3-319-49314-5 |pages=271–272 |url=https://books.google.com/books?id=wTmgDQAAQBAJ}}</ref>

{{math theorem|name=Theorem|note=|style=|math_statement=A function <math>f : A \subseteq \R \to \R</math> is continuous at <math>x_0</math> if and only if it is [[sequentially continuous]] at that point.
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{{collapse top|title=Proof|left=true}}
''Proof.'' Assume that <math>f : A \subseteq \R \to \R</math> is continuous at <math>x_0</math> (in the sense of [[(ε, δ)-definition of limit#Continuity|<math>\epsilon-\delta</math> continuity]]). Let <math>\left(x_n\right)_{n\geq1}</math> be a sequence converging at <math>x_0</math> (such a sequence always exists, for example, <math>x_n = x, \text{ for all } n</math>); since <math>f</math> is continuous at <math>x_0</math>
<math display=block>\forall \epsilon > 0\, \exists \delta_{\epsilon} > 0 : 0 < |x-x_0| < \delta_{\epsilon} \implies |f(x)-f(x_0)| < \epsilon.\quad (*)</math>
For any such <math>\delta_{\epsilon}</math> we can find a natural number <math>\nu_{\epsilon} > 0</math> such that for all <math>n > \nu_{\epsilon},</math>
<math display=block>|x_n-x_0| < \delta_{\epsilon},</math>
since <math>\left(x_n\right)</math> converges at <math>x_0</math>; combining this with <math>(*)</math> we obtain
<math display=block>\forall \epsilon > 0 \,\exists \nu_{\epsilon} > 0 : \forall n > \nu_{\epsilon} \quad |f(x_n)-f(x_0)| < \epsilon.</math>
Assume on the contrary that <math>f</math> is sequentially continuous and proceed by contradiction: suppose <math>f</math> is not continuous at <math>x_0</math>
<math display=block>\exists \epsilon > 0 : \forall \delta_{\epsilon} > 0,\,\exists x_{\delta_{\epsilon}}: 0 < |x_{\delta_{\epsilon}}-x_0| < \delta_\epsilon \implies |f(x_{\delta_{\epsilon}})-f(x_0)| > \epsilon</math>
then we can take <math>\delta_{\epsilon}=1/n,\,\forall n > 0</math> and call the corresponding point <math>x_{\delta_{\epsilon}} =: x_n</math>: in this way we have defined a sequence <math>(x_n)_{n\geq1}</math> such that
<math display=block>\forall n > 0 \quad |x_n-x_0| < \frac{1}{n},\quad |f(x_n)-f(x_0)| > \epsilon</math>
by construction <math>x_n \to x_0</math> but <math>f(x_n) \not\to f(x_0)</math>, which contradicts the hypothesis of sequential continuity. <math>\blacksquare</math>
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