Prove that if a vector space $V$ has a basis of $n$ vectors, then every basis of $V$ must consist of exactly $n$ vectors? The proof in my textbook says:
Let $B_1$ be a basis of $ n $ vectors and $ B_2$ be any other basis of $V$.
(a)Because $ B_1$ is a basis and $B_2 $ is linearly independent, $B_2 $ has no more than n vectors.
(b)Because B$_2$ is a basis and $ B_1$ is linearly independent, $B_2$ has at least $n$ vectors. So $B_2$ has exactly $n $ vectors.
I understand (a) because I understand that if a vector space $V$ has a basis $ B = {b_1,...,b_n} $ then any set in $ V$ containing more than $n$ vectors must be linearly dependent.
But I don't understand the logic of (b). 
SOS!
 A: Let $A$ be a basis with $m$ elements and $B$ be a basis with $n$ elements.  Since both of them are basis of the vector space they have two properties i.e both of them span the vector space and both of them are linearly independent.  Take $A$ and use the property that it spans the vector space and take $B$ with the property that it is linearly independent. Now try to see that the cardinality of a spanning set is always $\geq$ to the cardinality of a spanning set. So from here you get $m\geq n$. Now reverse the roles of $A$ and $B$ to get $m\leq n$. 
A: The logic goes somehow like this: suppose $B_1$ has less elements than $B_2$. Lets say $B_1=\{b_1,\ldots,b_m\}$ and $B_2=\{v_1,\ldots,v_n\}$ and $m<n$. Now you can replace $m$ elements in $B_2$ by $B_1$ and still have a set of linearly independent set of vectors. Say $B'=\{b_1,\ldots,b_m,v_{m+1},\ldots,v_n\}$. 
But now since $B_1$ was a basis you have that $\sum a_i b_i=v_{m+1}$ for some $a_i$ in your field. But this means
$$ \sum_{i=1}^m a_i b_i -v_{m+1}=0, $$
contradicting that $B'$ is linearly independent.
A: $a.~~$ $B_2~$ has no more than cardinality $~B_1~$ vectors.
$b.~~$ $B_1~$ has no more than  cardinality $~B_2~$ vectors.
Same logic when you annihilate $~n~$.
A: We can label the two bases whichever way round we like, so let's label them so that $|B_1| \le |B_2|$.
$B_1$ is a basis, so if $|B_2|$ were strictly greater than $|B_1|$ then $B_2$ would not be linearly independent. But $B_2$ is also a basis, so it must be linearly independent.
$\Rightarrow |B_1| = |B_2|$
A: Let $B_{1}$ and $B_{2}$ two basis for the vector space $V$ suchs that $|B_{1}| = n$, $|B_{2}| = m$ with $m < n$ then $V \simeq \mathbb{R}^{n}$ and $V \simeq \mathbb{R}^{m}$ so by transitivity $\mathbb{R}^{n} \simeq \mathbb{R}^{m}$ which implies that $n = m$ but this is a contradiction.
