Difference between revisions of "Continuity definitions are equivalent"
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| + | ==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> | ||
| − | {{Theorem|Topology}} | + | ==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> | ||
| + | |||
| + | ''Note:'' we have now shown that <math>\forall\epsilon>0\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 a\in f^{-1}(B_\epsilon(f(x)))\text{ which then }\implies f(a)\in B_\epsilon(f(x))</math> | ||
| + | |||
| + | Or just <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 Of|Topology}} | ||
Latest revision as of 07:20, 27 April 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
- [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 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]
Note: we have now shown that [math]\forall\epsilon>0\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 a\in f^{-1}(B_\epsilon(f(x)))\text{ which then }\implies f(a)\in B_\epsilon(f(x))[/math]
Or just [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.
