Subspace of Euclidean Space is Closed

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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, $V \subseteq \ker A$.

$\Box$


$\ker A \subseteq V$

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:

\(\ds \mathbf 0\) \(=\) \(\ds A \mathbf x\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} \ds \mathbf u_{k + 1}^\intercal \sum_{i \mathop = 1}^n \alpha_i \mathbf u_i \\ \vdots \\ \ds \mathbf u_n^\intercal \sum_{i \mathop = 1}^n \alpha_i \mathbf u_i \end {bmatrix}\)
\(\ds \) \(=\) \(\ds \begin {bmatrix} \alpha_{k + 1} \\ \vdots \\ \alpha_n \end {bmatrix}\) Definition of Orthonormal Basis of Vector Space

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:

$\ker A \subseteq V$

$\Box$


By definition of set equality:

$V = \ker A$

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

$\blacksquare$


Sources

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