Difference between revisions of "Continuity definitions are equivalent"

From Maths
Jump to: navigation, search
m
m
Line 1: Line 1:
{{Todo|Theorem B16 in Introduction to topological manifolds is a good guide to structure}}
 
  
 +
==Statement==
 +
The definitions of continuity for a function <math>f:(X,d)\rightarrow(Y,d')</math> from one [[Metric space|metric space]] to another is the same as <math>f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})</math> being continuous (where the topologies are those [[Topology induced by a metric|induced by the metric]] are the same, that is
 +
# <math>\forall a\in X\forall\epsilon>0\exists\delta>0:x\in B_\delta(a)\implies f(x)\in B_\epsilon(f(a))</math>
 +
# <math>\forall V\in\mathcal{K}:f^{-1}(V)\in\mathcal{J}</math>
 +
 +
==Proof==
 +
===<math>\implies</math>===
 +
Suppose <math>f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})</math> is continuous.
 +
 +
Let <math>V\in\mathcal{K}</math> - that is <math>V</math> is [[Open set|open]] within <math>Y</math>
 +
 +
Let <math>x\in f^{-1}(V)</math> be given.
 +
 +
Then because <math>V</math> is open, <math>\exists\epsilon>0</math> such that <math>B_\epsilon(f(x))\subset V</math> (note that <math>f(x)\in V</math> by definition of where we choose x from).
 +
 +
But by continuity of <math>f</math> we know that <math>\exists\delta>0:a\in B_\delta(x)\implies f(a)\in B_\epsilon(f(x))\subset V</math>
 +
 +
Thus <math>B_\delta(x)\subset f^{-1}(V)</math> (as for all <math>a</math> in the ball, the thing <math>f</math> maps it to is in the ball of radius <math>\epsilon</math> about <math>f(x)</math>).
 +
 +
Since <math>x</math> was arbitrary we have <math>\forall x\in f^{-1}(V)\exists\text{an open ball containing x}\subset f^{-1}(V)</math>, thus <math>f^{-1}(V)</math> is open.
 +
 +
===<math>\impliedby</math>===
 +
Choose any <math>x\in X</math>
 +
 +
Let <math>\epsilon>0</math> be given.
 +
 +
As <math>B_\epsilon(f(x))</math> is an open set, the hypothesis implies that <math>f^{-1}(B_\epsilon(f(x)))</math> is open in <math>X</math>
 +
 +
Since <math>x\in f^{-1}(B_\epsilon(f(x)))</math> and <math>f^{-1}(B_\epsilon(f(x)))</math> is open, it is a neighborhood to all of its points, that means
 +
 +
<math>\exists\delta>0:B_\delta(x)\subset f^{-1}(B_\epsilon(f(x)))</math>
 +
 +
Using the [[Implies and subset relation|implies and subset relation]] we see <math>a\in B_\delta(x)\implies f(a)\in B_\epsilon(f(x)))</math>
 +
 +
Thus it is continuous at <math>x</math>, since <math>x</math> was arbitrary, it is continuous.
  
 
{{Theorem|Topology}}
 
{{Theorem|Topology}}

Revision as of 16:13, 13 February 2015

Statement

The definitions of continuity for a function [math]f:(X,d)\rightarrow(Y,d')[/math] from one metric space to another is the same as [math]f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})[/math] being continuous (where the topologies are those induced by the metric are the same, that is

  1. [math]\forall a\in X\forall\epsilon>0\exists\delta>0:x\in B_\delta(a)\implies f(x)\in B_\epsilon(f(a))[/math]
  2. [math]\forall V\in\mathcal{K}:f^{-1}(V)\in\mathcal{J}[/math]

Proof

[math]\implies[/math]

Suppose [math]f:(X,\mathcal{J})\rightarrow(Y,\mathcal{K})[/math] is continuous.

Let [math]V\in\mathcal{K}[/math] - that is [math]V[/math] is open within [math]Y[/math]

Let [math]x\in f^{-1}(V)[/math] be given.

Then because [math]V[/math] is open, [math]\exists\epsilon>0[/math] such that [math]B_\epsilon(f(x))\subset V[/math] (note that [math]f(x)\in V[/math] by definition of where we choose x from).

But by continuity of [math]f[/math] we know that [math]\exists\delta>0:a\in B_\delta(x)\implies f(a)\in B_\epsilon(f(x))\subset V[/math]

Thus [math]B_\delta(x)\subset f^{-1}(V)[/math] (as for all [math]a[/math] in the ball, the thing [math]f[/math] maps it to is in the ball of radius [math]\epsilon[/math] about [math]f(x)[/math]).

Since [math]x[/math] was arbitrary we have [math]\forall x\in f^{-1}(V)\exists\text{an open ball containing x}\subset f^{-1}(V)[/math], thus [math]f^{-1}(V)[/math] is open.

[math]\impliedby[/math]

Choose any [math]x\in X[/math]

Let [math]\epsilon>0[/math] be given.

As [math]B_\epsilon(f(x))[/math] is an open set, the hypothesis implies that [math]f^{-1}(B_\epsilon(f(x)))[/math] is open in [math]X[/math]

Since [math]x\in f^{-1}(B_\epsilon(f(x)))[/math] and [math]f^{-1}(B_\epsilon(f(x)))[/math] is open, it is a neighborhood to all of its points, that means

[math]\exists\delta>0:B_\delta(x)\subset f^{-1}(B_\epsilon(f(x)))[/math]

Using the implies and subset relation we see [math]a\in B_\delta(x)\implies f(a)\in B_\epsilon(f(x)))[/math]

Thus it is continuous at [math]x[/math], since [math]x[/math] was arbitrary, it is continuous.

Topology