Subspace of Euclidean Space is Closed

Theorem
Let $\R^n$ be a Euclidean space.

Every subspace of $\R^n$ is closed in $\R^n$.

Proof
First we note that from Euclidean Space is Normed Vector Space, $\R^n$ has a norm $\norm {\, \cdot \,}$.

Let $V$ be a subspace of $\R^n$.

Let $\set {\mathbf v_1, \ldots, \mathbf v_k}$ be a basis for $V$.

Extend this to basis $\set {\mathbf v_1, \ldots, \mathbf v_k, \mathbf v_{k + 1}, \ldots, \mathbf v_n}$ for $\R^n$.

By using Gram-Schmidt procedure, there exists a set of orthonormal vectors $\set {\mathbf u_1, \ldots, \mathbf u_n}$ such that:


 * $\forall k \in \N_{> 0} : k \le n : \map \span {\mathbf v_1, \ldots, \mathbf v_k} = \map \span {\mathbf u_1, \ldots, \mathbf u_k}$

where $\span$ stands for the linear span.

Let $A \in \R^{\paren {n - k} \times n}$ be a matrix as follows:


 * $A = \begin {bmatrix} {\mathbf u_{k + 1} }^\intercal \\ \vdots \\ {\mathbf u_n}^\intercal \end {bmatrix}$

where $\mathbf u^\intercal$ denotes the transpose of $\mathbf u$ when expressed as a column matrix.

$V \subseteq \ker A$
By definition of orthonormality of $\set {\mathbf u_i}$ it follows that:


 * $\forall i \in \N_{> 0} : i \le k : A \mathbf u_i = 0$.

Hence, any linear combination of $\mathbf u_1, \ldots, \mathbf u_k$ lies in the kernel of $A$.

That is, $V \subseteq \ker A$.

$\ker A \subseteq V$
Suppose $\mathbf x \in \ker A$.

Let $\ds \mathbf x = \sum_{i \mathop = 1}^n \alpha_i \mathbf u_i$, where $\set {\alpha_i}$ are scalars.

By definition of kernel:


 * $A \mathbf x = 0$.

Then:

Hence:


 * $\forall i \in \N_{\ge k + 1} : i \le n : \alpha_i = 0$

So:
 * $\ds \mathbf x = \sum_{i \mathop = 1}^k$

That is:
 * $\mathbf x \in V$

Therefore by definition of subset:


 * $\ker A \subseteq V$

By definition of set equality:


 * $V = \ker A$

By Kernel of Linear Transformation between Finite-Dimensional Normed Vector Spaces is Closed, $V$ is closed.