Can anyone give me an example of an bounded and linear operator $T:\ell^\infty\to \ell^\infty$ (the space of bounded sequences with the usual supnorm), such that T has dense range, but is not surjective?

Here is an answer to an easier but related question. Proposition. There is a one to one operator $T$ from $\ell_1(2^{\aleph_0})$ into $\ell_\infty$ that has dense range. Of course, such an operator cannot be surjective because $\ell_1(2^{\aleph_0})$ is not isomorphic to $\ell_\infty$. My proof of the Proposition uses an old result of Bill Davis and mine (Remark 4 in Davis, W. J.; Johnson, W. B. On the existence of fundamental and total bounded biorthogonal systems in Banach spaces. Studia Math. 45 (1973), 173–179): $\ell_\infty$ has a biorthogonal system $(x_\alpha,x_\alpha^*)_{\alpha<2^{\aleph_0}}$ with $\x_\alpha\=1$ and $\sup_\alpha \x_\alpha^*\<\infty$ such that the linear span of $(x_\alpha)$ is dense in $\ell_\infty$. To prove the Proposition, define $T$ to be the norm one linear extension of the map $e_\alpha \mapsto x_\alpha$, where $(e_\alpha)$ is the unit vector basis for $\ell_1(2^{\aleph_0})$. This mapping obviously has dense range and is one to one because every biorthogonal system is countably linearly independent. Here is a variation on the OP's question: Is there a one to one bounded linear operator from $\ell_\infty$ into itself that has dense range but is not surjective? The interest in the variation is that this question is easily seen to be equivalent to: Are there quasicomplementary copies of $\ell_\infty$ in $\ell_\infty$ that are not complementary? (Recall that two closed subspaces of a Banach space are said to be quasicomplementary if their sum is dense and their intersection is $\{0\}$.) 


I could prove that if $T$ has dense then $T$ is surjective, in the cases where $T=S^{*}+W$, $W$ is weakly compact and $S:l^1\rightarrow l^1$ or when $T$ has totally disconnected spectrum. 

