Consider a $n$-dim real vector space $A$ (say $\mathbb{R}^n$), and there exists a collection of $n$ vectors $\{v_1,...,v_n\}$ such that it spans a dense subset of $A$.
I am wondering if this collection of vectors forms a basis of $A$? (It seems it is true, but I cannot prove it)
Thanks in advance.
Really thanks @Guillermo García Sáez for pointing out the topology and @AnneBauval for providing answer!
It is indeed a basis since "in $\mathbb{R}^n$ with its usual topology, any subspace is closed".
$span\{v_1,...,v_n\}$ is a subspace and it is dense, so it in fact is the whole space $\mathbb{R}^n$.
===== Proof of the quoted statement =====
Suppose there is a subspace $S$ that is not closed, that means there exists a converging sequence of vectors in $S$ but the limit is not in $S$, i.e., $$\{u_k\}\subset S\to u, u\not\in S.$$ Since $S$ is a subspace (say $dim(S)=m\leq n$), we can write down the basis $\{b_1,...b_m\}$. Also note $u$ is linearly independent to $\{b_1,...b_m\}$, then $\{b_1,...b_m,u\}$ is linearly independent in $\mathbb{R}^n$.
Consider the subspace $S^*$ spanned by $\{b_1,...b_m,u\}$, and reconsider the above converging sequence $\{u_k\}\subset S^*$. Under the basis $\{b_1,...b_m,u\}$, $$u_k=(u_{k1},...,u_{km},0),$$ while, $$u=(0,...0,1).$$ Since $u_k\to u$, we have $u_{ki}\to0$ and $0\to1$ which is a contradiction.