Difference between revisions of "Continuity definitions are equivalent"

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

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]

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.