Given a metric space and a subset of , it can be easily proved that derived set (set of all limit points of ) is closed. Does this hold if is a general topological space? Thanks!
Anyway, you can see that the idea of a limit point carries over into a topological space right?
Oops, Plato beat me to it, and his explanation is better than mine...
Thank you Plato, but I cannot understand your post completely, do you mean: A is fixed, but in different topological spaces, its derived set A' is different, some are closed, the others are not, right? My question is: Is it possible that A' is always closed in all topological spaces? I can't prove it, nor can I find any counterexamples.
I found a clue in Sierpinski's "Introduction to general topology". It is asserted as a theorem that (See Fig 1 below). But the proof denpends on the hypothysis that the space be Hausdorff (actually axiom, the lowest level of separation axiom, on which the infinity property can be satisfied, see Fig 2 below). But my question is: Is A' closed in all topological spaces (here "all topological spaces" means, of course, all topological spaces containing A, sorry for omitting this qualifier in my last post), which do not necessarily satisfy the axiom?
Fig 1. extracted from Sierpinski's "Introduction to general topology", Page 29-30.
Fig 2. extracted from Munkres' "Topology" Page 99. (both books are available online from e.g. wikipedia)
Suppose that is a limit point of the derived set , i.e. .
Suppose that is not in the closure of .
For any open set if then .
But that means is a limit point of .
By the definition of limit point, .
So we have shown that any open set that contains also contains a point of that is different from .
By definition of limit point that shows that is a limit point of .
In other words we have shown .
contains all of its limit points. It is closed.