Understanding of Theorem of Basis Theorem. Let $V$ be a vector space and suppose that one basis has $n$ elements, and another basis has $m$ elements. Then $m=n$.
My question is that the theorem says basis is unique, doesn't it? So, how can we prove of the theorem, can you hint me?
 A: This is false, bases are NOT unique, and this can be easily disproved by a single counterexample. (Consider $P_1(\mathbb{R})$, any constant function is a basis for this). What is true is your first statement, that all bases for some vector space $V$ are of some unique cardinality. 
A: This statement only guarantees uniqueness of the SIZE of the basis, not the elements that belong to it. For instance $\{ (1,0), (0,1)\}$ forms a basis for $\mathbb{R}^2$, and so does $\{ (3,0), (0, 6.2659484)\}$. These are both linearly independent sets that span $\mathbb{R}^2$ as a vector space over $\mathbb{R}$, and the both have different elements. Yet they have the same number of elements in the basis, namely $2$.
The proof for this follows from what I've seen referred to as the "replacement theorem." 
A: As was pointed out above, the theorem only says that all bases of a vector space $V$ have the same size. To prove the theorem, let $V$ be a vector space. Now suppose $\beta_1 = \{x_1, \dots, x_n\}$ and $\beta_2 = \{y_1, \dots, y_m\}$. We use the fact that if $A,B$ are subsets of a vector space with $A$ linearly independent and $B$ spanning the vector space, then $|A| \leq |B|$. Now since $\beta_1$ is linearly independent and $\beta_2$ spans $V$, we have that $|\beta_1| \leq |\beta_2|$. Now we have that $\beta_2$ is linearly independent and $\beta_1$ spans $V$, so $|\beta_2| \leq |\beta_1|$. Thus, $|\beta_1| = |\beta_2|$. Notice this proof works for bases that contain both finitely and infinitely many vectors.
