TITLE: Is the closedness of the image of a Fredholm operator implied by the finiteness of the codimension of its image?
QUESTION [19 upvotes]: Let $X$ and $Y$ be Banach spaces. A bounded operator $T\colon X\to Y$ is called Fredholm iff

The dimension of $\ker(T)$ is finite,
The codimension of the image $\mathrm{im}(T)$ is finite,
The image $\mathrm{im}(T)$ is closed in $Y$.

Question: Is the third condition redundant?
Some lecture notes I'm working through claim that the third condition follows from the second one together with the open mapping theorem. I've checked some books on functional-analysis without finding a proof of this.

REPLY [18 votes]: Assume that $T$ is injective, because if it is not we know that $\textrm{ker}(T)$ is closed so we can replace $T$ by the induced map from $X/\textrm{ker}(T)$ (which is a Banach space).
Now define $T' := X \oplus C \to Y$ by $T'(x, c) = T(x) + c$ where $C$ is a closed complement for the range of $T$.
This $T'$ is clearly bounded, linear and an isomorphism. So by the open mapping theorem $T'$ is open. Note that $\textrm{im}(T) = T'(X \oplus \{0\})$ which is closed.