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A compact topological group is a group $G$ equipped with a compact Hausdorff topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous.

The special unitary group isIt is closed in the finite-dimensional real vector space $M_n(\mathbb{C})$ because the equations $U^{\ast }U=I$ and $\det U=1$ are closed conditions. It is bounded because every column of a unitary matrix has norm one. The Heine-Borel theorem therefore makes it compact. Matrix multiplication is continuous, and inversion is continuous because $U^{-1}=U^{\ast }$. Hence $SU(n)$ is a compact topological group.

HereThe L1 norm and uniform norm satisfyso the identity from the $d_{\infty }$ metric to the $d_1$ metric is one-Lipschitz and therefore continuous. The two metrics do not induce the same topology on all of $C[0,1]$. Forone has $d_1(f_n,0)=1/(2n)\to 0$ but $d_{\infty }(f_n,0)=1$. Thus $d_1$ convergence need not imply uniform convergence.

A map $F:(X,d_X)\to (Y,d_Y)$ is Lipschitz continuous if there is $L<\infty$ such thatfor all $x,x^{\prime }\in X$. Evaluation at $a\in [0,1]$ is one-Lipschitz because

Choose distinct $a_0,\dots ,a_n\in [0,1]$ and defineThe Vandermonde determinant is nonzero, so $T$ is a bijection. It is one-Lipschitz for the two uniform metrics. If ${\ell }_i$ are the associated Lagrange cardinal polynomials, thenand henceThus $T^{-1}$ is also Lipschitz.

Let ${\hat{\mathcal{P}}}_n$ be the polynomials whose values lie in $[-1,1]$. Its image under $T$ lies in the bounded cube $[-1,1{]}^{n+1}$. It is closed: if $T(p_j)\to v$, then the Lipschitz inverse gives uniform convergence $p_j\to p=T^{-1}(v)$, and passing to the limit pointwise preserves $|p(x)|\le 1$. By the Heine-Borel theorem, $T({\hat{\mathcal{P}}}_n)$ is compact, and the continuous inverse carries compactness back to ${\hat{\mathcal{P}}}_n$. Therefore

Take the circle $S^1=\{(x,y)\in {\mathbb{R}}^2:x^2+y^2=1\}$ with its subspace topology. Write $W=(-1,0)$ for its west point and $E=(1,0)$ for its east point. Two coordinate charts obtained by stereographic projection areBoth take values in the entire real line. Geometrically, join $W$ to $(x,y)$ and intersect the resulting line with $x=0$: the height is $t$. Projecting from $E$ instead gives height $u$. Each projection omits its own starting point, but the other coordinate chart includes that point. Hence $U\cup V=S^1$.^[ref]

To verify the homeomorphisms, solve the displayed equations for $(x,y)$. The inverse functions areEach formula lies on the circle, avoids its excluded point, and is continuous for every real argument. Substitution into the projection formulas returns $t$ or $u$; substitution in the other direction returns $(x,y)$. Thus these are continuous bijections with continuous inverse functions, as required for coordinate charts. Since the Euclidean plane is Hausdorff and second countable, the subset $S^1$ with its subspace topology is too. This checks all three conditions for a topological manifold, rather than just drawing a smooth-looking curve.

On the overlap $U\cap V$, both $x=\pm 1$ are excluded, so $y\ne 0$. Using $y^2=1-x^2$, the topological transition map isFor example, $(x,y)=(3/5,4/5)$ has $t=1/2$ and $u=2$. The two connected components of the overlap, the upper and lower arcs, correspond to the positive and negative parts of the punctured real line. The values $t=0$ and $u=0$ represent $E$ and $W$, respectively, and belong to only one chart each. The reciprocal coordinate change and its inverse function are smooth functions away from zero, so this topological atlas also supplies a smooth atlas.

Why can neither chart be enlarged to a single coordinate chart on the whole circle? The circle is a closed set and a bounded set in the Euclidean plane, hence compact by the Heine-Borel theorem. A global coordinate chart would give a homeomorphism to a nonempty open set $O\subseteq \mathbb{R}$. Its image would be compact and therefore closed in the real line. By connectedness of the real line, this forces $O=\mathbb{R}$, which is not compact. This is the one-dimensional instance of no compact manifold has a single Euclidean chart. Locally the circle is indistinguishable from the real line; globally compactness obstructs a single chart.