Set closure and interior points

If $$A \subset X$$ where $$X$$ has a topology, is it generally true that the interior of $$A$$ is equal to the interior of the closure of $$A$$? This seems very reasonable to me, but probably only because I'm visualizing $$A$$ as a disc in the real plane. If it isn't true, what would be a counterexample?

thanks

HallsofIvy
Homework Helper
Take X to be the set of real numbers with the usual topology.

Let A be the set of all rational numbers between 0 and 1 (inclusive).

What is the interior of A? What is the closure of A?

What is the interior of the closure of A?

HallsofIvy said:
Take X to be the set of real numbers with the usual topology.

Let A be the set of all rational numbers between 0 and 1 (inclusive).

What is the interior of A? What is the closure of A?

What is the interior of the closure of A?

Okay this is a good example! Let me see if I have it straight. The interior of A is the largest open set contained in A, which would be the empty set in this case. I say this because the set has no interior points, or points for which a neighborhood can be found containing only points in A. The closure of A is the smallest closed set containing A, which is the interval [0, 1]. I say this because for every point in [0, 1], every neighborhood of that point contains points in A. But the interior of [0, 1] is (0, 1), which is not the empty set.

Did I understand correctly? I am a little foggy on the properties of real numbers, so I can't really back up my claims about the density of rationals at the moment.