I have very little knowledge of set theory and proof writing. This is a problem from “Introduction to Topology: Pure and Applied” by Colin Adams and Robert Franzosa.
INFORMATION GIVEN
Consider the set of rational numbers ℚ in ℝ with the standard topology. We claim that int(Q)=∅. Assume it is not, and suppose that U is a nonempty open set contained in ℚ. Let x be an element in U. Then there is an open interval (a, b) such that x ∈ (a, b) ⊂ U ⊂ ℚ. But between every pair of real numbers there is an irrational number. Thus every interval contains elements of ℝ − ℚ, and therefore so does U. This is a contradiction; hence int(Q)=∅.
While the interior of ℚ contains nothing, if we take the closure we get everything; specifically, cl(Q)=R
PROBLEM
Prove that Cl(ℚ) = ℝ in the standard topology on ℝ
THOUGHTS
I think I understand the int(Q) in R, standard topology example: the open sets in the standard topology of R are real number intervals (a, b). There are irrational numbers between a and b, but they are not open in R, so Q has no open sets in R, and therefore int(Q) = ∅.
Am I on the right track?
As for the problem, I'm still trying to wrap my head around closed sets and closure points. The closure of Q is the intersection of all closed sets containing Q... alright, I admit that I'm stuck.
As always, I appreciate any help.
The complement of the closure of $\Bbb Q$ is open and disjoint from $\Bbb Q$. If it is non-empty, it contains some open interval $(a,b)$ disjoint from $\Bbb Q$ ...