Closed kernel of a operator. If a linear operator on a Banach space is bijective (hence has a closed Kernel), will it imply it is continuous?
 A: No, this is false. Take the space $c_{00}$ of all bounded sequences $(a_n)_{n\in\mathbb N}$ of real numbers such that $a_n=0$ if $n\gg1$, endowed with the supremum metric. Define $A\colon c_{00}\longrightarrow c_{00}$ by $A\bigl((a_n)_{n\in\mathbb N}\bigr)=(na_n)_{n\in\mathbb N}$. Then $A$ is bijective, $\ker A=\{0\}$ (hence closed) and $A$ is discontinuous.
A: Any bijective (continuous, linear) operator on a Vector Space $X$, that is $T \in L(X)$, has kernel $\mathrm{ker}(T) = {0}$, which is obviously closed. And indeed, any linear continuous and bijective operator on a Banach space $X$ has bounded inverse $T^{-1} \in L(X)$ by the Bounded Inverse Theorem. The latter is an easy consequence of the Closed Graph Theorem.
The question is therefore, if you are working on a complete normed space (Banach space). If not, there are counterexamples, as for example seen by Jose Carlos' post.
A: If you assume Choice, the answer is no. Take any normed space (Banach or not), let $\{e_j\}_{j\in J}$ be a Hamel basis. Write $J=K\cup L$ with $K=\{k_n\}_{n\in\mathbb N}$. Now define a linear operator $T$ by 
$$
Te_{k_n}=ne_{k_n},\ \ \ \ Te_j=e_j,\ \ \ j\in L. 
$$
Then $T$ is bijective and unbounded. 
