How did you define a closed set in a topological set? Did you say it was the complement of an open set; or the closure of the set is the set itself?
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...
Let be a limit point of , an arbitrary neighborhood of . To prove that so that is closed, we must find an element in satisfying and . Since is a limit point of , there is in a point and . The statement can be proved if the space is metric, or Hausdorff, since we can find a smaller neighborhood around that is contained in and disjoint from ; the element is in this smaller neighborhood due to the fact that is a limit point of . But in a general space, I'm not clear if this smaller neighborhood can be found. Could you please give me a further detailed hint?
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)
I didn't think that, this holds only when axiom are satisfied, that's why I couldn't prove it. which post of mine gives u that impression? I repeatedly stress the question is in a general topological space.
OK, show me the proof. Note that axiom is not assumed.
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.