Does the rank-nullity theorem hold for infinite dimensional $V$? The rank nullity theorem states that for vector spaces $V$ and $W$ with $V$ finite dimensional, and $T: V \to W$ a linear map,
$$\dim V = \dim \ker T + \dim \operatorname{im} T.$$
Does this hold for infinite dimensional $V$?  According to this, the statement is false.  But according to this, page 4, the statement is still true.  I'm thoroughly confused.  
 A: The rank formula also holds in infinite dimensions, whether you use cardinal arithmetic for the dimensions, or just say $\infty + n = \infty$, and $\infty + \infty = \infty$ (but one should use cardinal arithmetic). The proof is basically the same as in the finite-dimensional case, you choose a basis $\mathcal{B}_1$ of $\ker T$, a basis $\mathcal{B}_2$ of $\operatorname{im} T$, let $\mathcal{B}_3$ consist of preimages of the elements of $\mathcal{B}_2$ (choose one preimage per element), then $\mathcal{B}_1 \cup \mathcal{B}_3$ is a basis of $V$. In the infinite-dimensional case, some form of the axiom of choice is required, while the finite-dimensional case can be proved without that.
A: Indeed there might be some problem, as $\dim V$ is undefined if $V$ is not finite dimensional. Unless one wants to identify $\dim V$ with the cardinality of any of its bases, that is. But then one should also define algebraic operations with cardinalities to make sense of the term $\dim \ker T + \dim \text{im} T$, and this is complicated. (I don't even know if it is possible in some meaningful way).
Instead, in the general case (both finite and infinite dimensional) you can use the following result: the map $\tilde T$, defined by
$$
\tilde{T}(v+\ker T):=T(v),
$$
is a vector space isomorphism of $V/\ker T$ onto $\text{im}(T)$. In the finite dimensional case, the rank-nullity theorem is an immediate corollary.
A: In the finite case, for an operator $\;T:V\to V\;$ , the dimension theorem says that
$$T\;\;\text{is surjective}\;\implies \dim\ker T=0\iff \ker T=\{0\}$$
But if we define
$$V:=\left\{\{x_n\}_{n\in\Bbb N}\subset\Bbb R\right\}\;,\;\;T:V\to V\;,\;\;T\{x_1,x_2,...\}:=\{x_2,x_3,...\}$$
then clearly $\;T\;$ is onto, yet
$$\ker T:=\left\{\{x_n\}\in V\;;\;x_i=0\;\;\forall\,i\ge 2\right\}\neq\{0\}$$
