Does Closed Graph imply Closed Range Suppose I have a bounded linear operator from a space $X$ to $Y$, both Banach.
I know that if D(T) and Ran(T) are closed, then the graph G(T) is closed in $X\times Y$.
However is the converse true ? Does G(T) closed imply that Ran(T) is closed ?
More generally, is $A, B$  closed if and only if $A\times B$ is closed ?
Thank you.

Every bounded operator has a closed graph, but not necessarily a closed range.  Consider for instance the inclusion operator from $C([0,1])$ to $L^1([0,1])$.  It is bounded, but its range is $C([0,1])$ which is not closed in $L^1$.
I don't see how your second question ("more generally") is related to this, but the answer is affirmative assuming $A,B$ are both nonempty.  Use the fact that the projection maps on a product space are continuous.
