Subspace is convex and closed set Let $V$ be a vector space. Would you help me to prove that if $A$ is a subspace of $V$ then $A$ is convex and closed set. 
I can prove that $A$ is convex (it's easy) and try to prove that $A$ is closed by showing that any sequences   {$x_n$} in $A$ that converge to $x$ implies $x\in A$ but don't have any idea.
Thanks.
 A: I expanded my comments into an answer.

In infinite dimensional normed linear spaces, subspaces are convex but not necessarily closed. 
Consider $l_\infty(\mathbb{R})$ which is the set of bounded sequences in $\mathbb{R}$ with the norm $|(a_n)_{n \in \omega}| = \sup a_n$. Note that the vector space structure is given by term by term addition and term scalar multiplication.
Then let $M$ be the subspace of sequences that have only finitely many nonzero terms. You can verify that $M$ is a subspace. 
The sequence $\alpha = (1, \frac{1}{2}, \frac{1}{3}, ..., \frac{1}{n}, ...)$ is a bounded sequence, hence is in $l_\infty(\mathbb{R})$. $\alpha \in \overline{M}$, the closure, since 
$|\alpha - (1, \frac{1}{2}, ..., \frac{1}{n}, 0, 0, 0, ...)|_\infty = \frac{1}{n + 1}$. 
So $\alpha$ is the limit of a sequence of elements from $M$. $\alpha$ is in the closure of $M$ but not an element of $M$. Hence $M$ is not closed. 
A: You mean $V$ is finite dimensional, no? We can then regard $V$ as the standard $\mathbb R^n$.
Consider a supplementary base $v_1,\ldots,v_n$ orthogonal to $A$. Then 
$$A=\{x \mid \forall i:\langle x,v_i \rangle =0\} = \bigcap_i {v_i}^\perp$$
 and, as $\langle\cdot,v_i\rangle$ is continuous, we have that ${v_i}^\perp$ is closed.
For infinite dimension, this is not true. Neither the topology is not uniquely determined, can be of many kind..
A: I suppose $V$ is a finite dimensional vector space over $\mathbb{R}$.
We can identify $V$ as $\mathbb{R}^n$.
Let $A$ be an $n - r$ dimensional subspace of $V$.
There is a bijective linear transformation $f\colon V \rightarrow V$ such that $f(A) = \{(x_1\dots,x_n) \in \mathbb{R}^n\colon x_1 =\cdots=x_r = 0\}$.
Hence $A$ is closed.
