# Redundant condition? What if the image is proper and dense?

Let $X$ and $Y$ be Banach spaces. A bounded operator $T:X\to Y$ is a Fredholm if

• 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 $Y$.

I've found notes saying that the third condition is redundant. But, what is "codimension" if the image is not closed? What if the image is proper and dense?

In the answer of this question Is the closedness of the image of a Fredholm operator implied by the finiteness of the codimension of its image?, a closed complement is used, but is there always a closed complement?

I mean, how can the third condition be redundant if it is necessary to the second one make sense?

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You can define the codimension of the image as $\dim Y/\operatorname{im}(T)$ (in general the codimension of a subspace $U$ in $X$ is $\dim X/U$). See also point 2. in the answer to: Question about Fredholm operator for a proof of the existence of a closed complement. – user45918 Oct 24 '12 at 20:02
Thanks for your comment, @bla. But $H^1(\Omega)$ is dense in $L^2(\Omega)$, what is the codimension of $H^1$? The answer you cited supposes finite codimension to begin with. – Otavio Kaminski Oct 25 '12 at 0:55
Octavio: The codimension of $H^1$ is the dimension of the vector space $L^2/H^1$. The dimension is the cardinality of any basis, whether finite or infinite. In the Fredholm case, finite codimension of the image is assumed as in your second bullet point. Point 3. of the answer bla links to answers your question. – Jonas Meyer Oct 25 '12 at 6:59
Thanks for your comment, @JonasMeyer. My question was "how many dimensions does $L^2/H^1$ have?" (or geometrically: how many dimensions are missing in $H^1$ to complete $L^2$ since $H^1$ is dense in $L^2$?) – Otavio Kaminski Oct 25 '12 at 12:25
Since the embedding $H^1 \to L^2$ is continuous and the image is dense the codimension is certainly infinite (by what Jonas explained and the linked answer). Whether it is countable or uncountable, I don't know. I would be surprised if it were countable... – user46056 Oct 26 '12 at 0:38