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.

But $\mathbb R^3$ is not a plane.
It's a three-dimensional space.  And the open set $P$ is not  a circle.  It is a sphere.... a $3D$ sphere... and a sphere can't lie entirely in a plane.
Any open ball $B_r(w_1, w_2, 0) = \{(x,y,z)\in \mathbb R^3| d((w_1,w_2,0), (x,y,z)) < r\}$ will include the point $(w_1,w_2, \frac 12 r)$.  (because $d((w_1,w_2,0),(w_1,w_2,\frac 12r)) = \sqrt {(w_1-w_1)^2 + (w_2-w_2)^2 + (\frac 12 r- 0)^2} =\frac 12 r < r$).  And $(w_1, w_2, \frac 12 r) \not \in \{(x_1, x_2, 0)|x_1,x_2 \in \mathbb R\}$.
So no, a 2-dimensional plane existing in a 3-dimensional space is not all open set.
