As stated, I know of course weakly closed implies weakly sequentially closed, but is the converse also true? In other words, I want to know if in weak topology, a subset is weakly closed if and only if it is sequentially closed. aka they are equivalent definitions just analogous to the case in norm topology. I also know that it is not true to say sequentially closed imply closed in ALL topological spaces. But is this true specifically in weak topology? Answers appreciated.
Does weakly sequentially closed imply weakly closed
functional-analysisgeneral-topology
Best Answer
The converse is false, even in Hilbert space. See Example 3.33 in Bauschke-Combettes's Convex Analysis and Monotone Operator Theory in Hilbert Spaces (second edition), which is a more general version of @mechanodroid Example.
The converse is true in finite-dimensional Banach spaces or when the set considered is convex.