Determine closure and interior of $\mathbb{Q} \cap (0,1)$
Definitions: A set is open iff for every point $x \in A$ we have a neighbourhood such that U(x) $\subset A$. $cl(A):=$ smallest possible closed set which contains A, $int(A):=$ biggest possible open set which is a subset of $A$
What I already could show is that closure of the ratoinals is equals to the reals and interor of the rationals is the empty set. The closure of $(0,1)$ is $[0,1]$ and the interior of $(0,1)$ is the set itself. Thus we find interior of $\mathbb{Q} \cap (0,1)=$ $int(\mathbb{Q}) \cap int((0,1))=\emptyset$. But I can't apply the same trick for the closure. Any hints?
Best Answer
Hint: The closure includes every point you can reach to as the limit of sequences from your subset. What points can you approximate arbitrarily closely with a sequence of rationals between 0 and 1?