[Math] A subspace of Hilbert space

Question

How can I prove that a finite dimensional subspace of Hilbert space is closed?
I assume that there is a sequence in this subspace which is convergent and I want to show that this limit is actually in the same subspace any help?

Answer 1

Suppose $Y$ is a finite-dimensional subspace of a normed space $X$. All the norms in finite-dimensional spaces are equivalent, so if $e_i$ is a basis of $Y$ and $y = \sum y_i e_i$ is the decomposition of $y\in Y$, there are constants $C_1>0$ and $C_2>0$ such that $$ C_1\|y\|_X \le \max |y_i| \le C_2 \|y\|_X$$ Let a sequence $y^n \in Y$ converges to $x \in X$, the components of $y_i^n$ of $y^n$ on the basis $e_i$ satisfy for $n, m \in {\mathbb N}$ $$\max_i |y_i^n - y_i^m|\le C_2 \|y^n - y^m\|_X$$ hence they are a Cauchy sequence. Let $y_i = \lim_n y_i^n$ and $y = \sum y_i e_i$. We have $$C_1 \|y^n - y\|_X\le \max_i |y_i^n - y_i|\mathop{\longrightarrow}_\limits{n\to \infty} 0$$ Hence $x = y\in Y$.