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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? enter image description here

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    $\begingroup$ Yup, one could replace $f:X\to\mathbb R$ with $f:X\to\mathbb R^m$ and the Theorem is still true :) $\endgroup$ Oct 20, 2020 at 18:16
  • $\begingroup$ Hmm, that is curious then. Thanks for clarifying! $\endgroup$
    – EE18
    Oct 20, 2020 at 18:31
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    $\begingroup$ 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 $\mathbb R$. $\endgroup$ Oct 20, 2020 at 19:13
  • $\begingroup$ @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). $\endgroup$
    – EE18
    Oct 20, 2020 at 20:48
  • $\begingroup$ Check out my answer below $\endgroup$ Oct 20, 2020 at 23:01

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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.

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  • $\begingroup$ Super, thanks very much! $\endgroup$
    – EE18
    Oct 20, 2020 at 23:52

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