Editing Net (mathematics) (section)

=== Continuity ===

A function <math>f : X \to Y</math> between topological spaces is [[Continuous function (topology)|continuous]] at a point <math>x</math> if and only if for every net <math>x_\bull = \left(x_a\right)_{a \in A}</math> in the domain, <math>\lim_{} x_\bull \to x</math> in <math>X</math> implies <math>\lim{} f\left(x_\bull\right) \to f(x)</math> in <math>Y.</math>{{sfn|Willard|2004|p=75}} 
Briefly, a function <math>f : X \to Y</math> is continuous if and only if <math>x_\bull \to x</math> in <math>X</math> implies <math>f\left(x_\bull\right) \to f(x)</math> in <math>Y.</math> 
In general, this statement would not be true if the word "net" was replaced by "sequence"; that is, it is necessary to allow for directed sets other than just the natural numbers if <math>X</math> is not a [[first-countable space]] (or not a [[sequential space]]).

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(<math>\implies</math>) 
Let <math>f</math> be continuous at point <math>x,</math> and let <math>x_\bull = \left(x_a\right)_{a \in A}</math> be a net such that <math>\lim_{} x_\bull \to x.</math>
Then for every open neighborhood <math>U</math> of <math>f(x),</math> its preimage under <math>f,</math> <math>V := f^{-1}(U),</math> is a neighborhood of <math>x</math> (by the continuity of <math>f</math> at <math>x</math>).
Thus the [[Interior (topology)|interior]] of <math>V,</math> which is denoted by <math>\operatorname{int} V,</math> is an open neighborhood of <math>x,</math> and consequently <math>x_\bull</math> is eventually in <math>\operatorname{int} V.</math> Therefore <math>\left(f\left(x_a\right)\right)_{a \in A}</math> is eventually in <math>f(\operatorname{int} V)</math> and thus also eventually in <math>f(V)</math> which is a subset of <math>U.</math> Thus <math>\lim_{} \left(f\left(x_a\right)\right)_{a \in A} \to f(x),</math> and this direction is proven.

(<math>\Longleftarrow</math>) 
Let <math>x</math> be a point such that for every net <math>x_\bull = \left(x_a\right)_{a \in A}</math> such that <math>\lim_{} x_\bull \to x,</math> <math>\lim_{} \left(f\left(x_a\right)\right)_{a \in A} \to f(x).</math> Now suppose that <math>f</math> is not continuous at <math>x.</math>
Then there is a [[Neighbourhood (mathematics)|neighborhood]] <math>U</math> of <math>f(x)</math> whose preimage under <math>f,</math> <math>V,</math> is not a neighborhood of <math>x.</math> Because <math>f(x) \in U,</math> necessarily <math>x \in V.</math> Now the set of open neighborhoods of <math>x</math> with the [[Subset|containment]] preorder is a [[directed set]] (since the intersection of every two such neighborhoods is an open neighborhood of <math>x</math> as well).

We construct a net <math>x_\bull = \left(x_a\right)_{a \in A}</math> such that for every open neighborhood of <math>x</math> whose index is <math>a,</math> <math>x_a</math> is a point in this neighborhood that is not in <math>V</math>; that there is always such a point follows from the fact that no open neighborhood of <math>x</math> is included in <math>V</math> (because by assumption, <math>V</math> is not a neighborhood of <math>x</math>).
It follows that <math>f\left(x_a\right)</math> is not in <math>U.</math>

Now, for every open neighborhood <math>W</math> of <math>x,</math> this neighborhood is a member of the directed set whose index we denote <math>a_0.</math> For every <math>b \geq a_0,</math> the member of the directed set whose index is <math>b</math> is contained within <math>W</math>; therefore <math>x_b \in W.</math> Thus <math>\lim_{} x_\bull \to x.</math> and by our assumption <math>\lim_{} \left(f\left(x_a\right)\right)_{a \in A} \to f(x).</math>
But <math>\operatorname{int} U</math> is an open neighborhood of <math>f(x)</math> and thus <math>f\left(x_a\right)</math> is eventually in <math>\operatorname{int} U</math> and therefore also in <math>U,</math> in contradiction to <math>f\left(x_a\right)</math> not being in <math>U</math> for every <math>a.</math>
This is a contradiction so <math>f</math> must be continuous at <math>x.</math> This completes the proof. 
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