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.
 A: 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.
A: Thanks to Nate for his answers, I think I understand it now. I am summarizing it here for the benefit of other readers.
Firstly, $D(T)$ closed, $\mathrm{Ran}(T)$ closed implies $D(T) \times \mathrm{Ran}(T)$ is closed and vice versa provided the sets are non-empty.
However, the graph of the operator $G(T) = \{(x, Tx)| x \in D(T)\}$ being closed does NOT imply that $\mathrm{Ran}(T)$ is closed.
Finally, the book "Functional Analysis" by Cloud & Vorowich has an error on page -154. 
They state that $G(T) = D(T) \times \mathrm{Ran}(T)$. This is false. $G(T)$ is in fact a subset of $D(T) \times \mathrm{Ran}(T)$
