Difference between revisions of "Continuous map"

Revision as of 00:28, 13 February 2015 (view source) Alec (Talk | contribs) (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...")		Revision as of 15:41, 13 February 2015 (view source) Alec (Talk | contribs) m Newer edit →
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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}}	+	==Equivalence of definitions==
		+	[[Continuity definitions are equivalent]]

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Revision as of 15:41, 13 February 2015

First form

The first form:

$$f:A\rightarrow B$$ is continuous at $$a$$ if:
$$\forall\epsilon>0\exists\delta>0:|x-a|<\delta\implies|f(x)-f(a)|<\epsilon$$ (note the implicit $$\forall x\in A$$ )

Second form

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

$$f:(A,d)\rightarrow(B,d')$$ is continuous at $$a$$ if:
$$\forall\epsilon>0\exists\delta>0:d(x,a)<\delta\implies d'(f(x),f(a))<\epsilon$$
$$\forall\epsilon>0\exists\delta>0:x\in B_\delta(a)\implies f(x)\in B_\epsilon(f(a))$$

In both cases the implicit

$$\forall x$$ is present. Basic type inference (the $$B_\epsilon(f(a))$$ is a ball about $$f(a)\in B$$ thus it is a ball in $$B$$ using the metric $$d'$$ )

Third form

The most general form, continuity between topologies

$$f:(A,\mathcal{J})\rightarrow(B,\mathcal{K})$$ is continuous if
$$\forall U\in\mathcal{K}\ f^{-1}(U)\in\mathcal{J}$$ - that is the pre-image of all open sets in $$(A,\mathcal{J})$$ is open.

Equivalence of definitions

Continuity definitions are equivalent