Prove $O(n)$ is compact

Question

I have to prove $O(n)$ is compact, I know if I can prove it bounded and closed in $\mathbb{R^{n\times n}}$, I will be done. But how to check boundedness and closed ness. For closedness I would like to show that all its limit points are inside $O(n)$, but what are the limit points of $O(n)$ and for boundedness I have no insight. Plesae tell me easiest possible ways without going into higher manifolds theory and all as I am a beginner in subject. Thanks!

Answer 1

The rows of a matrix representation of an element of $O(n)$ must form an orthonormal basis of $\mathbb{R}^n$. In particular, all entries are bounded in absolute value by $1$. To show it is closed, note that it is the solution set of a finite system of polynomial equations, corresponding to the condition that $AA^T=I$ for all $A\in O(n)$. It is the intersection of the inverse images of a point (the point $c$ such that $p_i(x_1,x_2,\ldots,x_{n^2})=c$ for one of the equations in the system) for each polynomial in the system, which is closed.

Answer 2

We can use the fact that if a function is continuous, the preimage of closed sets are also closed sets. For $u,v\in \mathbb{R^n}$, define $f_{u,v}:\mathbb{R^{(n^2)}}\rightarrow \mathbb{R^n}\times\mathbb{R^n}$, $f_{u,v}(A)=(Au,Av)$, then define $g:\mathbb{R^n}\times\mathbb{R^n}\rightarrow\mathbb{R}$ be the inner product of $n$-dimensional vectors. It is easy to prove that $h_{u,v}:=g\circ f_{u,v}$ is continuous, so $S_{u,v}:=h_{u,v}^{-1}(\{g(u,v)\})$ is closed, and $O(n) = \cap_{u,v\in \mathbb{R^n}}S_{u,v}$ is closed. Finally, the entries are bounded, so are the matrices.