Is irrational number ( $\mathbb{Q}^c$ ) are baires space ? yes/no
i got the links Irrational number and Baire space
But i didn't understand the answer properly
My Proof : By the Baire Category Theorem, a space that is homeomorphic to a complete metric space must be a Baire Space: the intersection of a countable family of open dense sets must be dense. But $\mathbb{Q^c}$ does not have this property, because it is uncountable and no point is isolated: for each $p\in\mathbb{Q^c}$, let $C_{p }= \mathbb{Q^c}\setminus\{p\}$. This is open (since $\{p\}$ is closed in $\mathbb{R}$, hence in the induced topology of $\mathbb{Q^c}$) and dense, since every open ball with center in $\mathbb{p}$ intersects $C_p$. But $$\bigcap_{p\in\mathbb{Q^c}} C_p = \emptyset$$ is not dense. Hence $\mathbb{Q^c}$ is not a Baire space.