I need to prove the following :
Interior of a Plane in $\mathbb{R^3}$ , under $d_2$ metric is empty .
($d((u_1,u_2,u_3), (v_1,v_2,v_3)) = \sqrt{(u_1 – v_1)^2 +(u_2 -v_2)^2 + (u_3 – v_3)^2}$)
Please check whether my proof is correct or not.
A plane in $\mathbb{R^3}$ is given by : $ax + by + cz = d$ where $a,b,c \in \mathbb{R}$
Let $U =\{ax + by + cz = d|a,b,c \in \mathbb{R}, and (x,y,z) \in \mathbb{R^3}\}$
Assume that Interior of $U$ is non empty ,say $(x_0 ,y_0,z_0) \in \text{Int(U)}$
there exist some $r >o $ such that open ball $B(t,r) \subseteq U$ where $t =(x_0 ,y_0,z_0)$
Clearly, $(x_0 +r/2,y_0,z_0) \in B(t,r)$
So, we have $a(x_0 + r/2) + by_0 + cz_0 = d $
and $ax_0 + by_0 + cz_0 = d$ , subtracting these two I get :
$a(r/2) = 0$, or $a =0$
Similarly I take $(x_0,y_0+r/2,z_0) \in B(t,r)$ $(x_0,y_0,z_0+r/2) \in B(t,r)$ to get $b =c =0$
So,if Interior of $U$ is non – empty the plane reduces to a point $\{0\}$, and this set is closed with empty interior.
Is my proof correct ? Please tell if there are any errors in my arguments.
Thank you.
Best Answer
It is almost correct, but there is a problem at the end. After having prove that $a=b=c=0$, you wrote that ”the plane reduces to the pont $\{0\}$“. Not quite. The conclusion is that the plane is defined by the expression $0=d$. But this is either the empty set (if $d\ne0$) or the whole space (if $d=0$). Since you are assuming that you are working with a plan, you have reached a contradiction.
Or you can say from the start that a plane is a subset of $\Bbb R^3$ of the form$$\{(x,y,z)\in\Bbb R^3\mid ax+by+cz=d\},$$where $a,b,c,d\in\Bbb R$ and the numbers $a$, $b$, and $c$ are not all equal to $0$ (that's the natural deffinition, in my opinion). This way, you get a contradiction in a slightly different way.