Proof of whether or not a set P is an open set Let $P=\{(x_1, x_2,0) :x_1, x_2 \in \mathbb{R}\}$ be the $x_1$-$x_2$ plane in $\mathbb{R}^3$. Is $P$ an open subset of $\mathbb{R}^3$?
I know that $P$ is an open set if for each points $x_1, x_2$ in $P$, there exists a radius $r>0$ such that $B_r(x) \subseteq P$. I think that $P$ is an open set because it is just a circle in the $\mathbb{R}^3$ plane. This is assuming the $0$ doesn't change in the definition of open considering it is not a variable. I am not sure how to prove the set is open from here.

Hint: it isn't open.
Negate the definition of an open set. Choose the origin (edit: tbc, the origin in $\mathbb{R}^3$). Let $r>0$ be given. Show that there exists a point within distance $r$ of the origin, yet which isn't in $P$. Conclude that the origin is not an interior point of $P$.
