Why is the closed $\operatorname{im} T$ in the definition of Fredholm Operator redundant? In this Wikipedia link, it's stated that the condition of $\operatorname{im} T$ being closed in the definition of Fredholm Operator is redundant. Why is that?
Edit: Most proofs use the existence of a closed $N$ such that $T:X\rightarrow Y$, $Y=imT\oplus N$. However, why does such a $N$ exist?
 A: The reference given on the wiki page is to [Abramovich/Aliprantis, Lemma 4.38]:
Any [linear] operator $T : X \to Y$ between Banach spaces with a finite defect has a closed range and satisfies [a dimension formula]. 
The proof refers to Corollary 2.17:
Let $T : X \to Y$ be a bounded [linear] operator between two Banach spaces. If the quotient vector space $Y / R(T)$ is finite dimensional, then the range of $T$ is closed. 
This is a corollary to Theorem 2.16:
A bounded [linear] operator $T : X \to Y$ between two Banach spaces has a closed range if and only if there exists a closed subspace $Z \subset Y$ such that $R(T) \cap Z = \{ 0 \}$ and the vector subspace $V := R(T) \oplus Z$ is closed in $Y$. 
Proof of the "if" part: 


*

*The vector space $V / Z$ with the quotient norm is a Banach space.

*$J : R(T) \to V / Z$, $J(y) := y + Z$ is a surjective linear isomorphism.

*With the notation $\|y\|_T := \|y\| + \|T^{-1} y\|$ for $y \in R(T)$, the latter norm being the quotient norm on $T / N(T)$, we have $\| J(y) \| \leq \| y \| \leq \| y \|_T$, so the operator $J$ is continuous, and, being one-to-one and onto, Theorem 2.5 below gives that $J$ is bounded below. 

*Thus the norms $\|\cdot\|$ and $\|\cdot\|_T$ are equivalent on $R(T)$. Since $R(T)$ is closed for $\|\cdot\|_T$, it is also closed for $\|\cdot\|$.


Theorem 2.5:
A continuous operator $T : X \to Y$ between Banach spaces is bounded below if and only if $T$ is one-to-one and has a closed range. 
The proof is not difficult and uses the Open Mapping Theorem at one point.
[Probably there is a more concise way to answer you question, but that, it seems to me, is how the book proceeds.]
