What shown below is a reference from "Analysis on manifolds" by James R. Munkres
First of all I desire discuss the compactness of $\Delta$: infact strangerly I proved the compactness of $\Delta$ in the following way. So we remember that if $Y$ is compact and if $Z\subseteq Y$is closed then $Z$ is compact; moreover if $Z$ is hausdorff separable then $\Delta Z:=\{(z,z):z\in Z\}$ is closed in $Z\times Z$; and finally if $Z\subseteq Y$ is compact and if $S\subseteq Z$ is compact in $Z$ then it is compact too in $Y$. So we observe that the compact $X$ is hausdorff separable, since $\Bbb{R}^m$ is hausdorff separable and since the hausdorff separability is hereditary; moreover $X\times X$ is compact and hausdorff separable, since the compactness and the hausdorff separability are multiplicative properties. So for what previously we observed, we can claim that $\Delta$ is compact in $\Bbb{R}^{2m}$. So is what I observed correct?
Now I desire to discuss the continuity of $g$. First of all we remember that $\Bbb{R}^n$ is a topological vector space thus the vector sum $s$ is continuous. So we define the function $\phi:X\times X\rightarrow\Bbb{R}^n\times\Bbb{R}^n$ through the condiction
$$
\phi(x,y):=\big(f(x),-f(y)\big)
$$
for any $x,y\in X$ and so we observe that $g\equiv ||\cdot||\circ s\circ\phi$. So we remember that the norm $||\cdot||$ is continuous (here the proof) thus if we prove that $\phi$ is too continuous then $g$ will be continuous, since the composition of continuous functions is too continuous. So let's start to prove the continuity of $\phi$. Clearly for the associativity of product topology $\Bbb{R}^n\times\Bbb{R}^n$ is homeomorphic to $\Bbb{R}^{2n}$ and so for $i=1,…,2n$ we can define $\pi_i\circ\phi$. So we observe that
$$
\pi_i\circ\phi=\begin{cases}f_i,\text{ if }i\le n\\ -f_i,\text{ otherwise}\end{cases}
$$
and so for the universal mapping theorem for products we can claim that $\phi$ is continuous and so the statement holds. So is what here I observed correct?
Could someone help me, please?
Best Answer
What you’ve done is correct, but it’s unnecessarily complicated. For instance, to show that $\Delta$ is compact you could argue in the same style but more simply that $X\times X$ is compact, because it’s the Cartesian product of two compact spaces, and $\Delta$ is a closed subset of $X\times X$ (and therefore compact) because $X$ is Hausdorff.
To show that $g$ is continuous, you need only note that $d(x,y)=\|x-y\|$ is a metric1 on $\Bbb R^n\times\Bbb R^n$, so it is continuous, and that the map
$$X\times X\to\Bbb R^n\times\Bbb R^n:\langle x,y\rangle\mapsto\langle f(x),f(y)\rangle$$
is continuous because it is continuous in each factor. (I don’t know whether you’ve already proved this result, but it’s standard and very easy. This map is the diagonal product of the map $f$ with itself, sometimes denoted by $f\Delta f$. See, for instance, Definition $13$ in this PDF.) Then $g$ is simply the composition of these two continuous maps: $g=d\circ(f\Delta f)$.
1 Specifically, the Euclidean metric, but that doesn’t matter.