Is $I:=\{a\in \text{End}_D(V):\text{dim}_D(aV)<\infty\}$ the only proper ideal of $\text{End}_D(V)$? Let $D$ be a division ring and $V$ a countably infinite dimensional vector space over $D$.  It is claimed that the ideal $$I:=\{a\in \text{End}_D(V):\text{dim}_D(aV)<\infty\}$$ is the only proper ideal of $\text{End}_D(V)$.   Suppose that I define $$J:=\langle \text{End}_D(V)\setminus I\rangle$$ as another ideal generated by elements of $\text{End}_D(V)$ not in $I$.  
Question:  
(1) Is $J=\text{End}_D(V)$?  
(2) Is $J$ another proper ideal of $\text{End}_D(V)$ such that $J=\{0\}$?
My reasoning is that $J=\text{End}_D(V)$ because I read some where and it was being claimed that $I$ is the only proper ideal of $\text{End}_D(V)$.  But then, I do not have reasons for why $I$ is the only proper ideal of $\text{End}_D(V)$.  
 A: Not sure exactly which question you care about, so it may or may not be helpful to note that you can find $J$ without knowing that $I$ is the only proper ideal:
Of course if $V$ has finite dimension then $I$ is not a proper ideal in the first place; $End_D(V)\setminus I=\emptyset$. Otoh if $V$ has infinite dimension then the identity map is an element of $End_D(V)\setminus I$...
A: I think the lemma you're fishing for is this:

Let $\phi,\psi\in End(V_D)\setminus I$.  Then there exists $x,y\in End(V_D)$ such that $x\phi y=\psi$. In particular, all elements out side of $I$ generate the same ideal as the identity element: $End(V_D)$.

Proof:  first, since $Im(\phi)$ and $Im(\psi)$ have the same dimension, there must obviously be a transformation $x$ such that $Im(x\phi)=Im(\psi)$ ($x$ can be an isomorphism between the two images.)
Now, fix a basis for $Im(\psi)$, say $\{v_i\mid i\in \mathbb N\}$. Now, select $w_i\in \psi^{-1}(v_i):=\{v\in V\mid \psi(v)=v_i\}$ for every $i$. It is an easy exercise to show $\{w_i\mid i\in \mathbb N\}$ is a linearly independent set, and so we can extend it to a basis $\mathscr B$.  Again, select $w'_i\in (x\phi)^{-1}(v_i)$ for every $i$. Let $y$ be the map such that $y(w_i)=w'_i$ and which is zero on the rest of $\mathscr{B}$.
By definition, then, $x\phi y=\psi$, since they are equal on the basis $\mathscr{B}$.

More generally, for any infinite cardinality $\kappa$, you can show that the proper ideals of the endomorphism ring of a $\kappa$ dimensional vector space correspond to the infinite cardinals $\lambda$ such that $\aleph_0\leq \lambda < \kappa$ using the same argument. They are linearly ordered, and you can classify them as endomorphisms whose images have dimension less or equal to one of those cardinalities.
You can see why the cardinality comes into play: if the image of $\psi$ is of higher dimension than $\phi$, then you are not going to be able to increase the dimension of $x\phi y$ using any $x$ or $y$.  You can only maintain or decrease the dimension.
