# Proof from Hubbard: Why restrict to the reals only?

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? • Yup, one could replace $f:X\to\mathbb R$ with $f:X\to\mathbb R^m$ and the Theorem is still true :) Oct 20, 2020 at 18:16
• Hmm, that is curious then. Thanks for clarifying!
– EE18
Oct 20, 2020 at 18:31
• Not only you can replace $\mathbb R$ by $\mathbb R^m$ in the proof, but also the case with codomain $\mathbb R^m$ follows from the case with codomain $\mathbb R$. Oct 20, 2020 at 19:13
• @AndréPorto can you elaborate on that? I can see how I could argue identically to the proof above to get the case for $\mathbb{R}^m$, but I don't see how it follows immediately (I only have theorems for coordinate-wise convergence being equivalent to convergence in the sense of sequences).
– EE18
Oct 20, 2020 at 20:48
• Check out my answer below Oct 20, 2020 at 23:01

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.