Compact subset of a finite dimensional inner product space

Question

In a finite dimensional inner product space, every bounded and closed set is compact.

My understanding:

Let $X$ be a finite say $n$ dimensional inner product space. every $x \in X$ can be expressed as

$x = \sum_{i=1}^{n}\alpha_i v_i$ with $a_i$ scalars and $v_i$ are the elements from the basis set.

To show that a bounded and closed subset $S$ of $X$ is compact, one needs to show that every sequence $(s_n)$ in $S$ has a convergent sub-sequence $(s_{n_{k}})$ in $S$.

I am not sure how to show this.

Answer 1

The map $\sum \alpha_i v_i \to (a_1,a_2,...,a_n)$ is an isomorhpism. Since linear maps are continuous on finite dimensional spaces this is a homeomorphism. So it preserves boundedness, closedness and compactnes. In the usual topology of $\mathbb R^{n}$/$\mathbb C^{n}$ any closed and bounded set is compact so the same is true in the given space.

Answer 2

Since you have a finite-dimensional inner product space, you can always find an orthonormal basis (via Gram-Schmidt). Let $(e_1, \ldots, e_n)$ be such a basis for an inner product space $X$, and suppose $K \subseteq X$ is closed and bounded by $M$ (i.e. $v \in K \implies \|v\| \le M$).

Consider the linear maps $\langle \cdot, e_i \rangle$. Using Cauchy-Schwarz, you can show that these linear maps are continuous. Moreover, by the same inequality, if $x \in K$, then $$|\langle x, e_i \rangle| \le \|x\| \cdot \|e_i\| = \|x\| \le M.$$ Consider a sequence $(x_i) \in K$. The sequence $\langle x_i, e_1 \rangle$ is contained in the compact interval $[-M, M]$, and so there must be a convergent subsequence, say, $(x_{i_j})$. Then $\langle x_{i_j}, e_2\rangle$ is a sequence contained in $[-M, M]$, so another convergent (sub-)subsequence. A third (sub-sub-)subsequence guarantees that $\langle x_i, e_3\rangle$ converges, etc. After $n$ iterations (a sub-sub-...-subsequence), you'll have a sequence (call it $(y_j)$) such that $\langle y_j, e_i \rangle$ converges to some $a_i \in [-M, M]$ for all $i = 1, \ldots, n$ (as $j \to \infty$). This sequence will be a subsequence of $(x_i)$.

So, with this in mind, let's prove that $y_j \to a_1 e_1 + \ldots + a_n e_n =: y$. By a standard result, we have, for each $j$, $$y_j = \langle y_j, e_1 \rangle e_1 + \ldots + \langle y_j, e_n \rangle e_n.$$ Thus, by Pythagoras's theorem, $$\|y_j - y\|^2 = |\langle y_j, e_1 \rangle - a_1|^2 + \ldots + |\langle y_j, e_n \rangle - a_n|^2,$$ which is the sum of $n$ sequences that all converge to $0$, hence $\|y_j - y\| \to 0$, i.e $y_j \to y$. That is, we started with an arbitrary sequence $(x_i)$, formed a subsequence $(y_j)$, and constructed a limit for the sequence. How do we know that $y \in K$? Because $K$ is closed!