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.
 A: $(0, 0, 0) \in P$, but for any $r > 0$, $B_r((0, 0, 0))$ contains the point $(0, 0, r/2)$ which is not in $P$. So $P$ is not open. (Your description of what it means for $P$ to be open is nearly right, but you only need to talk about one $x$, not $x_1$ and $x_2$.)
Another way of seeing this is to note that $P$ is the inverse image of $\{0\}$ under the continuous function from $\Bbb{R}^3$ to $\Bbb{R}$ which maps a point $(x, y, z)$ to $z$. As this function is continuous, the inverse image $P$ of the closed set $\{0\}$ must also be closed. But the only subsets of $\Bbb{R}^3$ that are both open and closed are $\emptyset$ and $\Bbb{R}^3$, and, as $P$ is neither of these, it is not open. This approach may involve ideas about continuity and connectedness that you haven't studied yet, but when you have learnt those ideas, it can give you an "at a glance" way of seeing whether a set defined by a system of one or more constraints is open or closed. (The constraint in your example is $z = 0$. If you changed it to $z > 0$ you would get an open set.)
A: 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 $P$.
A: 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.
