To show that the special orthogonal group ${\rm SO}(n,\mathbb R)$, carrying the subspace topology induced by ${\rm Mat}_{n}(\mathbb R) \cong {\mathbb R}^{n}$, is compact many proofs use the Heine-Borel theorem.
As an alternative proof, does it suffice to observe that the continuous, real-valued function $\det: {\rm SO}(n, \mathbb R) \rightarrow \mathbb R$ is bounded on ${\rm SO}(n, \mathbb R) $ (because $\det(A) = 1$ for all $A \in {\rm SO}(n, \mathbb R)$) and to deduce that this is equivalent to compactness of ${\rm SO}(n, \mathbb R)$ (by another theorem)? If not, why?
Best Answer
No. For this to work you would need something like the determinant to be a proper map, that is a map such that the inverse image of every compact set is again compact.
Not every continuous map is proper, for example consider multiplication $ℝ × ℝ → ℝ$. The inverse image of $\{1\}$ is the hyperbola, which is unbounded and so certainly not proper.
As Omnomnomnom mentioned in the comments, the special linear group $\operatorname{SL}_n (ℝ)$ is an unbounded inverse image of $\{1\}$ under the determinant, so the determinant is not proper. In fact, the set $$\Big\{\smash{\big[\begin{smallmatrix}x & 0\\ 0 & y\end{smallmatrix}}\big]; x, y ∈ ℝ,~xy = 1\Big\}$$ yields an embedded hyperbola in the special linear group.