Basis of a vector space is a maximal linearly-independent set? 
If $V$ is a vector space of finite dimension over $F$, then a basis of
  $V$ is a maximal, linearly independent set in $V$.

Is this conjecture true? If so, how to prove it?
 A: Suppose $B=\{b_1,\dots,b_n\}$ is a basis for $V$ with $\dim V=n$. Then $B$ is necessarily a linearly independent set. Let $v_{n+1}\in V$ and consider $B'=B\cup\{v_{n+1}\}$. Since $v_{n+1}\in V$, it can be written as a linear combination of $v_1,\dots,v_n$. But that means $v_{n+1}$ is linearly dependent on $B$. So $B\cup\{v_{n+1}\}$ is not a linearly independent set and hence cannot be a basis for $V$. 
So yes, your statement is true.
A: Yes, it is true. Suppose $B=\{v_1,\ldots,v_n\}$ is a basis for $V$. Let $u\in V-B$ be any other vector. Then, by definition of a basis, $B$ spans $V$, so
$$u=a_1v_1+\cdots+ a_nv_n$$
for some $a_i\in F$ and hence $B\cup\{u\}$ is linearly dependent. So $B$ is a linearly independent set such that adding any vector to it yields a linearly dependent set. That is the definition of a maximal linearly independent set.
A: A basis is a spanning set, so can express any (other) vector as a linear combination. Rephrasing: to a basis, there is no such thing as another linearly independent element. 

The theorem would appear to be true of any vector space, finite-dimensional or not. 
