Unitary Group Compact

The map $U\mapsto U^{\ast }U$ is continuous because each entry is a finite sum of products of real-coordinate polynomial expressions from [F2]. The singleton $\{I_d\}$ is closed in the finite-dimensional Euclidean matrix space, so its inverse image $U(d)$ is closed. If $U=(u_{ij})\in U(d)$, each column has squared Euclidean norm ${\sum }_i\mathrm{\mid }{u}_{ij}{\mathrm{\mid }}^2=1$; hence every real and imaginary entry lies in $[-1,1]$. Thus $U(d)$ is bounded in ${\mathbb{R}}^{2d^2}$.

By Heine–Borel and the agreement of metric and topological compactness in [F1], the closed bounded set $U(d)$ from step 1.1 is compact in its Euclidean subspace topology. Distinct points of $U(d)$ have disjoint sufficiently small Euclidean balls, so its subspace topology is Hausdorff by [F5]. It is nonempty because $I_d\in U(d)$.

By [F3], each $U\in U(d)$ has inverse $U^{\ast }$, so the defining set is closed under inverses. If $U,V\in U(d)$, then $(UV{)}^{\ast }(UV)=V^{\ast }(U^{\ast }U)V=V^{\ast }V=I_d$, so it is closed under multiplication; associativity and the identity follow from [F4]. Matrix multiplication is continuous by the finite coordinate formulas of step 1.1, and inversion is $U\mapsto U^{\ast }$, which is continuous by [F2]. Thus the group with its subspace topology is a topological group by [F5].