Task:Continuity types

Overview

The continuity page is missing (links) to some information it ought to provide. This task page (the first task page) is a task to be done which holds the information required and can be completed in steps at a later date.

Somewhere the following should happen:

1. [ilmath]f:\mathbb{R}\rightarrow\mathbb{R} [/ilmath] is continuous at [ilmath]x\in\mathbb{R} [/ilmath] if:
   • [ilmath]\forall \epsilon>0\exists\delta>0\forall a\in\mathbb{R}[\vert a-x\vert < \delta\implies\vert f(a)-f(x)\vert < \epsilon][/ilmath] - and a discussion of why this is intuitive, and how to generalise it to higher dimensions (sticking to [ilmath]\mathbb{R}^n[/ilmath] of course)
2. Metric space version of 1) (for [ilmath](X,d_1)[/ilmath] and [ilmath](Y,d_2)[/ilmath] being metric spaces, [ilmath]f:X\rightarrow Y[/ilmath] a map and [ilmath]x\in X[/ilmath], [ilmath]f[/ilmath] is continuous at [ilmath]x[/ilmath] if):
   • [ilmath]\forall \epsilon>0\exists\delta>0\forall a\in X[d_1(x,a)<\delta\implies d_2(f(x),f(a))<\epsilon][/ilmath]
3. [ilmath]f:X\rightarrow Y[/ilmath] (for topological spaces [ilmath](X,\mathcal{ J })[/ilmath] and [ilmath](Y,\mathcal{ K })[/ilmath]) we have "topological continuity at a point":
   • [ilmath]\forall N\in\mathcal{N}_Y(f(x))[f^{-1}(N)\in\mathcal{N}_X(x)][/ilmath] - where [ilmath]\mathcal{N}_Z(p)[/ilmath] is the set of all neighbourhoods of [ilmath]p\in Z[/ilmath] for a topological space [ilmath](Z,\mathcal{ Z })[/ilmath].
4. (CLAIM: 1) - Warning:Unproven claim - 1) and 3) are in some sense equivalent. Obviously definition 1 doesn't work for a general topology, so we can only say equivalent on metric spaces or [ilmath]1)\implies 3)[/ilmath] - or something. As this is should be a fairly friendly to newcomers page, this should be covered carefully

Proof of claims

Claim 1

Let [ilmath](X,d_1)[/ilmath] and [ilmath](Y,d_2)[/ilmath] be metric spaces, then for a map [ilmath]f:X\rightarrow Y[/ilmath] and a point [ilmath]x\in X[/ilmath] we have:

• [ilmath]\Big(\forall \epsilon>0\exists\delta>0\forall a\in X[d_1(x,a)<\delta\implies d_2(f(x),f(a))<\epsilon]\Big)\iff\Big(\forall N\in\mathcal{N}_Y(f(x))[f^{-1}(N)\in\mathcal{N}_X(x)]\Big)[/ilmath]

Proof:

1. [ilmath]\Big(\forall \epsilon>0\exists\delta>0\forall a\in X[d_1(x,a)<\delta\implies d_2(f(x),f(a))<\epsilon]\Big)\implies\Big(\forall N\in\mathcal{N}_Y(f(x))[f^{-1}(N)\in\mathcal{N}_X(x)]\Big)[/ilmath]
   • Let [ilmath]N\in\mathcal{N}_Y(f(x))[/ilmath] be any neighbourhood of [ilmath]f(x)\in Y[/ilmath]. We must show [ilmath]f^{-1}(N)[/ilmath] is a neighbourhood of [ilmath]x\in X[/ilmath]
     • This means [ilmath]N[/ilmath] contains an open set containing [ilmath]f(x)[/ilmath]
     • This means there is an open ball of some radius [ilmath]r>0[/ilmath] centred at [ilmath]f(x)[/ilmath], [ilmath]B_r(f(x))[/ilmath] such that [ilmath]B_r(f(x))\subseteq\mathcal{O}\subseteq N[/ilmath] (where [ilmath]\mathcal{O} [/ilmath] is some open set to [ilmath]f(x)[/ilmath] the neighbourhood must have - because it's a neighbourhood)
       • By using [ilmath]r[/ilmath] for [ilmath]\epsilon[/ilmath] (as [ilmath]\forall \epsilon>0\exists\delta>0\forall a\in X[d_1(x,a)<\delta\implies d_2(f(x),f(a))<\epsilon][/ilmath] is true for all [ilmath]\epsilon>0[/ilmath]) we see (by hypothesis):
         • [ilmath]\exists\delta>0\forall a\in X[d_1(x,a)<\delta\implies d_2(f(x),f(a))<r][/ilmath]
       • Taking the [ilmath]\delta>0[/ilmath] we know exists we see that:
         • If [ilmath]a\in B_\delta(x)[/ilmath] then [ilmath]f(a)\in B_r(f(x))[/ilmath] (remember [ilmath]p\in B_t(q)\iff d(p,q)<t[/ilmath])
       • Now we have: [ilmath]a\in B_\delta(x)\implies f(a)\in B_r(f(x))\subseteq N[/ilmath]
       • It isn't hard to see that [ilmath]B_\delta(x)\in f^{-1}(N)[/ilmath].
       • But wait! [ilmath]B_\delta(x)[/ilmath] is a neighbourhood of [ilmath]x[/ilmath]!

That completes the proof. Just flesh this out for the final page.