Wednesday, 19 October 2016

continuity - Continuous map $mathbb{R}^nrightarrowmathbb{R}^n$




When we say some map $\phi=(\phi_1,\ldots,\phi_n)$ is a continuous map $\mathbb{R}^n\rightarrow\mathbb{R}^n$ we really mean that each component $\phi_i$ is continuous as a function $\mathbb{R}^n\rightarrow\mathbb{R}$ or do we mean something else? $\mathbb{R}^n$ is as a vector space, so we can also consider the distance between any two vectors $\|v-u\|$ and use the definition of continuity on $\mathbb{R}^n$ as a whole. Which approach is the "proper one"? I'm kind of confused.



Edit



I know now that there is theorem that says that component-wise continuity is equivalent to continuity. I will try to prove this assertion.



Let $\phi=(\phi_1,\ldots,\phi_n):\mathbb{R}^n\rightarrow\mathbb{R}^n.$



1) Let each of $\phi_i$ be continuous functions from $\mathbb{R}^n$ to $\mathbb{R}$. Let $\|\cdot \|$ denote euclidean norm in $\mathbb{R}^n$. From continuity we have

$$\forall i \quad \forall x \quad\forall\epsilon>0\quad \exists \delta >0\quad\forall y :\|x-y\|<\delta\implies|\phi_i(x)-\phi_i(y)|<\epsilon/\sqrt{n}$$


Answer



This is one of those cases where category theory clears up the mystery about products of sets with structure. Once the product in $\text Set$ of a collection of sets $\left \{ X_{\alpha } \right \}_{\alpha \in I}$ is obtained, using the universal mapping property (UMP), then if the individual $X_{\alpha }$ have structure---for example, if they have topologies or if they are groups-- then one can say exactly what the product $\mathit must$ be, in order for the UMP to be preserved.



Now, the fact, for example, that $\mathbb R^{\omega }$ is metrizable in the product topology but not in the box toplogy, tells us in some sense, that the product topology is the "right one" to use and the box topology is not if we want nice results; i.e. the UMP is what determines how we should view products. Of course, this does not mean that we can always get what we'd like: for example, if $I$ is uncountable, $\mathbb R^{I}$ is not metrizable in the product toplogy.



As for norms, they are all equivalent on $R^{n}$ for $n\in \mathbb N$, but not on infinite-dimensional spaces. For example, the $l_{p}$ spaces are all different with $l_{p}\subset l_{q}$ whenever $q>p$, the obvious example being the sequence $\left \{ \frac{1}{n} \right \}_{n\in N}$ which is in $l_{2}$ but not in $l_{1}$.


No comments:

Post a Comment

real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...