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? 

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.
