Proving the set $S=\left\{ \left(x,y\right);ax+by<c\right\} $ is open

Question

Prove the set $S=\left\{ \left(x,y\right);ax+by<c\right\} $ is open

My Approach I know there are many methods of proving this.But i find the method of proving every point to be interior point very fundamental.Please help me with this method only.

Let $A=\left(x_{o},y_{o}\right)$$\in S$ Now i need to prove that a Ball $B\left(A,r\right)$$\subset$$S$. I am unable to find any Ball.

Then i Saw Book's Approach It says

$\left(x_{o},y_{o}\right)$$\in S$ $\Longrightarrow$$ax_{o}+by_{o}<c$$\Longrightarrow$I Don't Understand how $ax_{o}+by_{o}<c$$\Longrightarrow$$\delta<\frac{|ax_{o}+by_{o}-c|}{\sqrt{a^{2}+b^{2}}}$

I know it is delta neighbourhood around $\left(x_{o},y_{o}\right)$.But how did they get this expression

and how can we say N$_{\delta}$$\left(A\right)$$\subset$S

Answer 1

Let $\varphi : \Bbb R^2 \to \Bbb R$ be given by $(x,y) \mapsto ax+by$. It is a linear combination of the two standard projection maps, so it is continuous, so the pre-image of any open set is an open set, and in particular $S = \varphi^{-1}[(-\infty,c)]$ is the preimage of the open interval $(-\infty,c)$, so it is open.