Difference between revisions of "Continuous map"

Revision as of 05:49, 15 February 2015

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

$\forall x\in A$ )

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

$d'$ )

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.

@@ Line 5: / Line 5: @@
 <math>f:A\rightarrow B</math> is continuous at <math>a</math> if:<br />
-<math>\forall\epsilon>0\exists\delta>0:|x-a|<\delta\implies|f(x)-f(a)|<\epsilon</math> (note the [[Implict qualifier|implicit <math>\forall x\in A</math>]])
+<math>\forall\epsilon>0\exists\delta>0:|x-a|<\delta\implies|f(x)-f(a)|<\epsilon</math> (note the [[Implicit qualifier|implicit <math>\forall x\in A</math>]])
 ==Second form==