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. $f:\mathbb{R}\rightarrow\mathbb{R}$ is continuous at $x\in\mathbb{R}$ if:
   • $\forall \epsilon>0\exists\delta>0\forall a\in\mathbb{R}[\vert a-x\vert < \delta\implies\vert f(a)-f(x)\vert < \epsilon]$ - and a discussion of why this is intuitive, and how to generalise it to higher dimensions (sticking to $\mathbb{R}^n$ of course)
2. Metric space version of 1) (for $(X,d_1)$ and $(Y,d_2)$ being metric spaces, $f:X\rightarrow Y$ a map and $x\in X$, $f$ is continuous at $x$ if):
   • $\forall \epsilon>0\exists\delta>0\forall a\in X[d_1(x,a)<\delta\implies d_2(f(x),f(a))<\epsilon]$
3. $f:X\rightarrow Y$ (for topological spaces $(X,\mathcal{ J })$ and $(Y,\mathcal{ K })$) we have "topological continuity at a point":
   • $\forall N\in\mathcal{N}_Y(f(x))[f^{-1}(N)\in\mathcal{N}_X(x)]$ - where $\mathcal{N}_Z(p)$ is the set of all neighbourhoods of $p\in Z$ for a topological space $(Z,\mathcal{ Z })$.
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 $1)\implies 3)$ - or something. As this is should be a fairly friendly to newcomers page, this should be covered carefully

Proof of claims

Claim 1

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

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

Proof:

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

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