Suppose V is a linear subspace of $\mathbb{R}^n$. How to show it's closed in $\mathbb{R}^n$

Question

We want to show that if $V\subseteq\mathbb{R}^n$ is a linear subspace of $\mathbb{R}^n$, then it's closed.

Let $V\subseteq\mathbb{R}^n$ arbitrary e suppose $V$ linear subspace of $\mathbb{R}^n$. As $V$ is a linear subspace of $\mathbb{R}^n$ so we have these properties:

$\textbf{0}\in V$

$x+y\in V$ for all $x,y \in V$

$\alpha x\in V$ for all $\alpha \in \mathbb{R}^n$ and $\alpha \in V$.

Suppose that $V$ is not closed. As $V$ is not closed, I know that exists a convergent sequence ($\textbf{x}_k$), with $\textbf{x}_k\in V$ for all $k\in\mathbb{N}$ such that $\lim\limits_{k\rightarrow\infty}\textbf{x}_k\notin V$. Let $(\textbf{x}_k)$ be a convergent sequence such that $\textbf{x}_k\in V$ for all $k\in\mathbb{N}$ and $\textbf{x}=\lim\limits_{k\rightarrow\infty}\textbf{x}_k\notin V$. Consider the following set $A_{\textbf{x}}=\left\lbrace\left\|\textbf{x}-\textbf{y}\right\|:\textbf{y}\in V\right\rbrace$. In fact, let $\varepsilon>0$ arbitrary. So $\textbf{x}_k\rightarrow\textbf{x}$, let $K^*\in\mathbb{N}$ such that $\left\|\textbf{x}_k-\textbf{x}\right\|<\varepsilon$ for $k\geq K^*$, therefore, in particular for $k=K^*$ we have $\left\|\textbf{x}_{K^*}-\textbf{x}\right\|<\varepsilon$ . But, $\textbf{x}_{K^*}\in V$, because $\textbf{x}_k\in V$ for all $k\in\mathbb{N}$ e $K^*\in\mathbb{N}$. Then, by definition of $A_{\textbf{x}}$ and by $\textbf{x}_{K^*}\in V$ we have that $||\textbf{x}_{K^*}-\textbf{x}||\in A_{\textbf{x}}$.Therefore, because $||\textbf{x}_{K^*}-\textbf{x}||\in A_{\textbf{x}}$, we have that $\inf A_{\textbf{x}}\leq \left\|\textbf{x}_{K^*}-\textbf{x}\right\|$ (infimum definition). Then, $\inf A_{\textbf{x}}\leq \left\|\textbf{x}_{K^*}-\textbf{x}\right\|$ and $\left\|\textbf{x}_{K^*}-\textbf{x}\right\|<\varepsilon$. From this,$\inf A_{\textbf{x}}<\varepsilon$. As $\varepsilon$ was arbitrary choosen, for all $\varepsilon>0$, $\inf A_{\textbf{x}}<\varepsilon$. Therefore we can $\inf A_{\textbf{x}}=0$.

After that I would like to use that exists a unique $\textbf{x}_*\in V$ such that $\left\|\textbf{x}_*-\textbf{x}\right\|=\inf A_{\textbf{x}}$, with this I found a contradiction with $\textbf{x}\notin V$. But, for showing this I need that V is closed. Is that other way to show it's closed without that assumption that V is closed or another way to answer the question without using continuity?

Thanks!

Answer 1

Let $(V,\|\cdot \|_V)$ a proper subspace. Suppose that $$V=Span\{v_1,...,v_m\},$$ where $m<n$. Complete it in a basis of $\mathbb R^n$, i.e. $$Span\{v_1,...,v_{m},v_{m+1},...,v_n\}=\mathbb R^n.$$

Notice that if $v\in V$, then for all $\varepsilon >0$ $$\mathcal B(v,\varepsilon )\cap V^c\neq \emptyset$$ since for example $v+\frac{\varepsilon }{2}v_n\in \mathcal B(v,\varepsilon )\setminus V$ for all $\varepsilon >0$.

I recall that $$\mathcal B(v,\varepsilon )=\{w\in\mathbb N\mid \|v-w\|_V<\varepsilon \}.$$

Answer 2

Let $V$ be a proper subspace of $\Bbb R^n$, denote by $\{e_1,e_2,\dots,e_n\}$ an orthonormal basis of $\Bbb R^n$ and define an linear map $f:\Bbb R^n\to \Bbb R^n$ by $f(x)=\langle x,\sum_{i=m+1}^ne_i\rangle$. Note that, $\ker f= V$, that is, $f^{-1}(0)=V$. Since $f$ is continuous, follows that $V$ is closed.