How to show that the vector subspaces of $\mathbb{R}^{n}$ are closed in $\mathbb{R}^{n}$? The vector subspaces of $\mathbb{R}^{n}$ are closed in $\mathbb{R}^{n}$.
How to show this? 
 A: Let $V\subseteq \mathbb{R}^n$ be a vector subspace of $\mathbb{R}^n$. Let's say it is of dimension $k$, where $0\leq k\leq n$. Let $\{v_1,\ldots,v_k\}$ be a basis for $V$. This basis for $V$ can be extended to a basis for $\mathbb{R}^n$, let's say
$$\{v_1,\ldots,v_k,w_1,\ldots,w_{n-k}\}.$$
Because this is a basis for $\mathbb{R}^n$, any element $y\in \mathbb{R}^n$ has a unique representation as a linear combination
$$y=a_1v_1+\cdots+a_kv_k+b_1w_1+\cdots+b_{n-k}w_{n-k},\qquad a_i,b_i\in\mathbb{R}.$$
We can define linear maps $\alpha_i:\mathbb{R}^n\to\mathbb{R}$ for $i=1,\ldots,k$, and linear maps $\beta_i:\mathbb{R}^n\to\mathbb{R}$ for $i=1,\ldots,n-k$ by
$$\alpha_i(y)=a_i,\quad \beta_i(y)=b_i$$
where these are the uniquely determined coefficients in the expression of $y$ in the basis we've chosen. Now prove that any linear map to $\mathbb{R}$ is continuous, and that therefore
$$V=\{y\in\mathbb{R}^n\mid \beta_1(y)=\cdots=\beta_{n-k}(y)=0\}=\bigcap_{i=1}^{n-k}\beta_i^{-1}(0)$$
is an intersection of closed sets, hence is closed.
A: Any subspace $E\mathbb \subset R^n$ is very canonically the common zero  set of the family of all (continuous !) linear forms $\sum a_ix_i$ vanishing on $E$. [In the symbolism of annihilator sets : $E=E^{\circ\circ}$]  
A: Every vector subspace $U$ of $\mathbb{R}^n$ has a basis consisting of $m \le n$ elements. Let this basis be $\{b_1,b_2, \ldots b_m \}$. Define the isomorphism $T: U \to \mathbb{R}^m$ by $T(b_1) = (1,0, \ldots 0)$. $T(b_2) = (0,1,\ldots 0)$ etc. So $U$ is isomorpic to $\mathbb{R}^m$ which is complete so $U$ is closed. 
Added:
This shows that U is closed in $\mathbb{R}^n$, because being closed is equivalent to the following: if $\{x_n\}_{n}$ is a sequence in $U$ and the sequence converges to $x \in \mathbb{R}^n$ then $x \in U$. Let $\{x_n\}_n$ be such a convergent sequence ($x_n \to x$). We have $\|T(x_n)−T(x_m)\|≤\|T\| \|x_n−x_m\|\to 0$. Since ${x_n}$ is convergent so also Cauchy. Also notice that the operator norm $\|T\|$ is finite as it can be expressed as a matrix. Since $\mathbb{R}^m$ is complete it follows that $T(x_n)$  converges to some $y \in mathbb{R}^m$. We have $T(x_n) \to y \in \mathbb{R}^m$, but then $T^{-1}(T(x_n)) = x_n$ converges to $T^{-1}(y) \in U$ (since $T^{-1}$ is continuous). But then $x = y \in U$ as limits are unique in $\mathbb{R}^n$. Hence the limit of the convergent sequence is in $U$, thus $U$ is a closed subspace of $\mathbb{R}^n$.  
A: Using incomplete basis theorem, every subspace $V$ of $\mathbb{R}^n$ is isometric to $\mathbb{R}^m$ where $m=\dim(V)$. Therefore, $V$ is complete and so closed in $V$.
A: Let $V\subseteq\Bbb R^n$, suppose that $\dim V=m$, then there is a basis $\{v_1,\ldots,v_n\}$ such that $v_1,\ldots,v_m$ is a basis for $V$. Let us write $k$ for $n-m$, note that $k\geq 0$.
Let $e_1,\ldots,e_k$ the standard basis for $\Bbb R^k$. Define the linear operator $T\colon\Bbb R^n\to\Bbb R^k$ as the unique operator for which: $$T(v_i)=\begin{cases}0 & 1\leq i\leq m\\ e_{i-m} & m<i\leq n\end{cases}$$
Since $T$ is continuous, as a linear map between finite dimensional spaces, $T^{-1}(0)=\ker T=V$ is closed as wanted.
