cardinality of any two bases are same How to prove that, any two bases of an infinite dimensional vector space have the same cardinality.
The proof for finite dimensional is easy.
But , I would like to see the most elementary and easy proof of this theorem.Most of the book solves this problem which unfortunately I don't understand.
Any help would be appreciated.
Thanks in advance.
 A: $\DeclareMathOperator{\Vect}{Vect}$
There is a somewhat easy proof assuming you are familiar with cardinals (and therefore with ordinals):
Let $E$ be a vector space over some field $k$. Let $\kappa$ be an infinite cardinal. Let $X$ be a generating subset for $E$ of cardinal $\kappa$, let $Y$ be a free subset of $E$. Let's prove by induction on $\kappa$ that $|Y| \leq |X|$.
To do this, assume that this is true for cardinals below $\kappa$, and fix a well-ordering $\leq_X$ of $X$ whose resulting order type is $\kappa$. Note that if $\kappa = \aleph_0$, this assumption is in fact a consequence of the result for finite families, so we can treat the base case along with the inductive step.
For $x \in X$, define $Y_x:= Y \cap \Vect(x' \in X, x' <_X x)$. Since any $\{x' \in X \ | \ x' <_X x\}$ has cardinal below $\kappa$ and $Y_x$ is a free subset of $\Vect(x' \in X, x' <_X x)$, by our hypothesis, we have $\forall x \in X, |Y_x| < \kappa$. 
$Y = \bigcup \limits_{x \in X} Y_x$ is thus a union indexed by $\kappa$ of sets of size $< \kappa$, so $|Y| \leq \kappa = |X|$.

You can see that this is not conceptually close to the proof in the finite case (which is arguably easy if one has never seen the argument!), but I actually don't know a proof which doesn't rely on the finite case.
I have put the three parts of the proof that use knowledge about cardinals on emphasis. I doubt you will benefit from studying this theorem without first trying to understand them. If you are mainly interested to the case when $\kappa = \aleph_0$, then those three points should seem more obvious.
