How do you prove that this set is open?

Question

$$a,b,c,d \in \mathbb{R} $$ $$ A = \{ (x,y,z)\in \mathbb{R^3} / ax + by + cz + d > 0 \} $$

I need to get this proof by definition, but I have ran into trouble trying to work with the difinition in $\mathbb{R^3}$, as I have no previous experience in that space.

Answer 1

For an arbitrary point $P(x_0, y_0,z_0 )$in $$A = \{ (x,y,z)\in \mathbb{R^3} / ax + by + cz + d > 0 \}$$

The open ball centered at P and radius $$ \delta = \frac {| ax_0+by_0 +cz_0-d|}{ 2\sqrt {a^2+b^2+c^2}}$$ is entirely in A.

Thus A is an open set.

Answer 2

Following the comment of @Mathematician 42, observe that $A=f^{-1}(0,\infty)$ where $f(x,y,z)=ax+by+cz+d.$ But $f$ is formed by addition of scalar multiples of three projection maps and a constant. Further, $(0,\infty)$ is an open set in $\Bbb R$. This means...

Answer 3

A set is open iff its complement is closed.

The complement of $$A = \{ (x,y,z)\in \mathbb{R^3} / ax + by + cz + d > 0 \}$$ is $$B = \{(x,y,z)\in \mathbb{R^3} : ax + by + cz + d \le 0 \}$$

Note that B is the space under and including the plane $$ ax + by + cz + d =0$$ which is its boundary, so B is closed.