Just a simple construction to illustrate Nate Eldredge's answer about functions with dense graphs. Pick any $mathbb{R}$-vector space E with a norm. On E, choose a non-continuous linear form $L: E to mathbb{R}$; now this can only be done if $dim(E)=infty$, of course.
Then, pick y such that $L(y)=1$, and let $T: E to E$ be defined by $Tx=x-L(x)y$. Then obviously T maps E onto the kernel of L; it is not difficult to prove that $ker (L)$ must be dense in E for any non-continuous L (the two conditions are even equivalent), and thus the graph of T must be dense in $E times E$.
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