Fredholm Index: Finite Corank $\Rightarrow$Closed Range

Question

Obviously closed subspaces turn quotient spaces into normed spaces rather than just merely vector spaces. However the dimension involved in Freholm's index are purely algebraic.

Why do we thus require the range to be closed?
Or is a subspace with finite codimension necessarily closed?

Moreover, what could happen if the range wouldn't be closed but would be of finite codimension though?

I'm thinking of some example like: $l^p_0\subsetneq l^p,1\leq p\leq\infty$

Answer 1

A bounded operator with finite cokernel has closed range (see Abramovich and Aliprantis, An Invitation to Operator Theory, Corollary 2.17).

Answer 2

You need the range to be closed in order to make sense of the cokernel. If $T:X \to Y$ then $$ \text{coker}(T) = Y / \text{range}(T).$$ In order for $Y / \text{range}(T)$ to be a Banach space you require $\text{range}(T)$ to be closed.