If then for some open set.
But any open set that contains x must also contain a point of A because x is in the closure of A.
Now if Q is any open set containing x then is also open and as such must contain a point of .
Can you finish now.
I think I solved it, but I did it differently, and I don't follow some of your explanation. But maybe I'm confused about something ... do elements of a set have to be either limit points or interior points? This seems to be implied in your explanation, but I thought this wasn't true? (for example the set of natural numbers in the usual topology of Reals).
As I understand it the closure is the union of a set and it's derived set:
So that your second statement about any open set containing x also containing another point of A doesn't make sense to me ... unless you didn't mean "another" point, but included the possibility of x itself?
I solved it like this. First I changed it to the equivalent:
If then ...
I just realized an error with what I was doing ... I wanted to write next:
or
But I see now I can't directly rule out or
And I'm concerned that my derived set equality above is not accurate, though it seemed straightforward a moment ago. I can at least show:
Though that goes the wrong direction, so I would need equality.
Well, now I've managed to thoroughly confuse myself, so I think I better take a closer look at your method.
I think I got this one now. Either x is in A or A´, if it is in A, then you get directly A\B, otherwise you get (A\B)´ (since there has to be another point of A\B in the open set) ... but it seems to be the introduction of Q was not necessary ... I don't see what this adds. The first open set seems to work fine by itself.