First if I am correct is assuming that is the closure operator, then I am puzzled by your confusion.
Here are some facts.
, and .
Because A is dense in X you also have .
Put all that together.
Hello everyone.
I am trying to prove that if is dense then for any open we have . I tried assuming not, and I arrive at the statement and I can't see why that would be a contradiction.
Does anyone know of a good proof of this? Also, is it necessary that the T be open?
No I did not. If T is open then the statement follows as a matter of definitions. If there is a t which is not in then realize that t is a limit point of A. But some open set O contains t and is a subset of T. Think about that contradiction.
You asked about T being open. Let . It is clear that the statement is false using this example.
I am sorry, but I still don't see it. Since is dense means that intersects every non-empty open set. In particular . I see this is the reason why it fails if is not open.
Also since is dense means that every point is a limit point of , so in particular is a limit point of . And since is open, we can find an open neighborhood of , , such that .
Now, you asked me to assume that and consider that open set defined above. Since is a limit point of implies that . But I still don't see the contradiction.