# Compact subset of a finite dimensional inner product space

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.

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!
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.