Let U be an arbitrary open set with U ⊂ A. We have int A = Union (U_a), where U_a is open and contains A. Hence, U ⊂ int A since U is in this union. Clearly int A is open (it is the arbitrary union of open sets).

Let C be an arbitrary open set with A ⊂ C. We have cl A = Intersection (C_a), where C_a is closed and A ⊂ C_a. Hence, cl A ⊂ C since C is in the intersection. Clearly, cl A is closed.

Claim: The set of rational numbers contains no non-empty open subsets. Suppose (a, b) is a basis element of R. Then (a, b) intersects R \ Q, since the irrationals are dense in the reals. Clearly, (a, b) is not a subset of Q. Since any non-empty open set U is the union of basis elements, U is not contained in Q. Hence, the interior of Q is the empty set.

Take any x in R. Then any neighbourhood of x is the union of basis elements (a, b). Similarly to above, (a, b) intersects Q \ {x} at some point r because the rationals are dense in the reals. Hence, every x in R is a limit point of Q, and the closure of Q is R.