Proof from Hubbard: Why restrict to the reals only?

Question

I'm reading through Hubbard and Hubbard's Vector Calculus, Linear Algebra, and Differential Forms, and in it the following theorem and proof are supplied. Now I followed the proof, but what I didn't understand was why the function $f$ was restricted to have the codomain $\mathbb{R}$. I didn't notice the hypothesis used anywhere and (if indeed the hypothesis is not needed) I can't see why the author wouldn't allow for a codomain of $\mathbb{R}^m$ more generally seeing as this is a book on multivariable calculus. The book is fantastic and it goes without saying that Hubbard is a far more capable mathematician than I am, so I'm wondering where I'm missing something here?

Answer 1

You can replace $\mathbb R$ by $\mathbb R^m$ in the proof, really by just replacing them, since the properties he used on the norm of $\mathbb R$ are the same on $\mathbb R^m$.

Also the case with codomain $\mathbb R^m$ follows imediatelly from the case with codomain $\mathbb R$. To see that, fix $f: X\subset\mathbb R^n \to \mathbb R^m$ and consider its coordinate functions $f_1, ..., f_m$. Each $f_i$ is a continuous function from the compact $X$ to $\mathbb R$, so by the result with codomain $\mathbb R$, each one of them is uniformly continuous.

Given $\varepsilon>0$, for each $f_i$, we may fix $\delta_i>0$ satisfying $$ |x-y|<\delta_i\ \Rightarrow\ |f_i(x)-f_i(y)|< \frac{\varepsilon}{\sqrt m}. $$ Put $\delta=\min\{\delta_1, ..., \delta_m\}>0$. Note that \begin{align} |x-y|<\delta & \Rightarrow\ |x-y|<\delta_i,\ \forall i\in\{1, ..., m\} \\ & \Rightarrow\ |f_i(x)-f_i(y)|< \frac{\varepsilon}{\sqrt m},\ \forall i\in\{1, ..., m\} \\ & \Rightarrow\ |f(x) - f(y)|= \sqrt{\sum_{i=1}^m |f_i(x)-f_i(y)|^2} < \sqrt{\sum_{i=1}^m \frac{\varepsilon^2}{m}} = \varepsilon, \end{align} so $f$ is uniformly continuous.