Correctness of Statements about Set A in Metric Space X

[Euge]

Furthermore, since $A$ has empty interior, $A$ cannot be open, making (c) the correct answer.

 

[I like Serena]

I do not see how $A$ has empty boundary. The closure of $A$ is $[0,1]$ since the irrationals are dense in the reals, and $A$ has empty interior since the rationals are dense in the reals. So wouldn't $\partial A$, being $\bar{A} - \operatorname{Int}(A)$, equal $[0,1]$?

The topology is $X=\mathbb R\setminus\mathbb Q$ with the usual metric.
So none of the rationals can be part of either the interior or the boundary.
So the closure of $A$ is $[0,1]\setminus \mathbb Q$, which is the same as the interior of $A$, making the boundary $\partial A$ empty, doesn't it?

 

[Euge]

Looking back at the original question, it appears that I overlooked that $A$ inherits the subspace topology relative to $X$, not to $\Bbb R$. My apologies. Certainly $A = (0,1) \cap X$, which is relatively open in $X$, and $X - A = (\Bbb R - \Bbb [0,1]) \cap X$, which is relatively open in $X$; thus, $A$ is both open and closed in $X$, so that $A$ has empty boundary.