Determine the interior and closure of the set $A=\mathbb{Q}\cap (0,1)$.
My approach: First note that the set of rationals $\mathbb{Q}$ and irrational $\mathbb{I}=\mathbb{R}\setminus\mathbb{Q}$ are dense in $\mathbb{R}$. So we have that, the interior of $A$ is empty, because if $x\in A$ is a rational number, then there exits an irrational number $y\in \mathbb{R}\setminus\mathbb{Q}$ such that the distance between them is so closed.
For the other hand, we have the closure of $A$ is $[0,1]$ since, if $x\in A$, then there exist rationals number $y$ arbitrarily close to $x$. For each $n\in\mathbb{N}$, choose $a_{n}\in A$ such that $|x-a_n|<\frac{1}{n}$. Then we have that $\lim_{n\to\infty}{a_{n}}=x$
Best Answer
Your Proof is fine!
Alternatively, $$(0,1)\cap \Bbb{Q} \subset \Bbb{Q} \Longrightarrow\text{Int}[(0,1)\cap \Bbb{Q}] \subset \text{Int}(\Bbb{Q})=\varnothing$$
For the second one , you have proved all points in $A$ is a limit point of $A$ . Which is obvious. Actually we have to prove all points in $[0,1]$ are limit points of $A$. So we pick an $x$ arbitrarily in $[0,1]$ and using density of rational to produce a sequence which converges to $x$