Let $A=\{(x,y)|x \ is\ an\ integer\}$. Then, is $A$ both open and closed in $\mathbb{R}^2$ in usual topology?
I think it it neither open nor closed, because it is a union of lines, which are nowhere dense in $\mathbb{R}^2$
On a similar note, is the plane $\mathbb{R}^2$ open in $\mathbb{R}^3$ with usual topology?
Again, I think no, because of the nowhere denseness of the set. Any hints. Thanks beforehand.
$A$ can't be both open and closed since $\mathbb{R}^{2}$ is connected with usual topology. In fact $A$ is closed (all lines parallel to $y-axis$ intersecting $x-axis$ at integer points), you can do this directly by looking at the complement of $A$ and proving that every point in complement has ball of sufficiently small radius.
Other way of proving is to look at the projection map (which is continuous ) $\pi : \mathbb{R}^{2} \rightarrow \mathbb{R}$ onto the first coordinate. Then $\pi^{-1}(\mathbb{Z}) = A$.