Difference between revisions of "Continuous map"

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(Created page with "{{Definition|Topology|Metric Space}} ==First form== The first form: <math>f:A\rightarrow B</math> is continuous at <math>a</math> if:<br /> <math>\forall\epsilon>0\exists\de...")
 
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<math>\forall U\in\mathcal{K}\ f^{-1}(U)\in\mathcal{J}</math> - that is the pre-image of all open sets in <math>(A,\mathcal{J})</math> is open.
 
<math>\forall U\in\mathcal{K}\ f^{-1}(U)\in\mathcal{J}</math> - that is the pre-image of all open sets in <math>(A,\mathcal{J})</math> is open.
  
{{Todo|The most important Theorem, that these two continuity definitions are the same}}
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==Equivalence of definitions==
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[[Continuity definitions are equivalent]]

Revision as of 15:41, 13 February 2015


First form

The first form:

f:AB is continuous at a if:
ϵ>0δ>0:|xa|<δ|f(x)f(a)|<ϵ (note the implicit xA)

Second form

Armed with the knowledge of what a metric space is (the notion of distance), you can extend this to the more general:

f:(A,d)(B,d) is continuous at a if:
ϵ>0δ>0:d(x,a)<δd(f(x),f(a))<ϵ
ϵ>0δ>0:xBδ(a)f(x)Bϵ(f(a))

In both cases the implicit x is present. Basic type inference (the Bϵ(f(a)) is a ball about f(a)B thus it is a ball in B using the metric d)

Third form

The most general form, continuity between topologies

f:(A,J)(B,K) is continuous if
UK f1(U)J - that is the pre-image of all open sets in (A,J) is open.

Equivalence of definitions

Continuity definitions are equivalent