My problem is the following:
I know that but I need to figure out & understand the ``how'' behind that answer, in other words, give a proof that A-closure is that set above. Thanks.
Do you understand that the closure of a set is the union of the set with all of its limit points?
If so then all we have to do is show that is the set of limit points of .
For any point and any
does the interior of the disk contain a point of
If so, is a limit point.
I partially understood what you just said. However, some of the things I am having trouble with are the notions of closure & limit points. I do understand the first statement, though. But some of the rest I am a bit vague on. Sorry, but I really am trying to understand these ideas as much as possible.
Have you graphed the function y= sin(1/x)? Close to x= 0, the graph oscillates infinitely many times between -1 and 1. For any point (0, y) with , there exist points on the graph arbitrarily close to (0, y). That's why any such point is in the closure of the graph.
In particular, given y between -1 and 1, there exist some z such that sin(z)= y. Then also, so that letting , goes to 0 as n goes to infinity and goes to y: the points converge to (0, y) so (0, y) is a limit point.