TITLE: Prove that open half planes are open sets
QUESTION [6 upvotes]: An open half plane is a subset of $\Bbb{R}^2$ in the form $\{(x, y)\in \Bbb{R}^2\vert \space Ax + By<C\}$ for some $A,C,B\in \Bbb{R}$ with either $A$ or $B$ nonzero. 
I need to prove that open half planes are open sets in the standard topology on $\Bbb{R}$. I understand conceptually what an open half plane is and I understand the concept of an open set in the standard topology. I'm struggling with the methodology of this proof.

REPLY [2 votes]: Let $X$ be a topological space and $U\subset X. $ If for every $p\in U$ there exists an open $V$ such that $p\in V\subset U$ then $U$ is open: 
(1). For each $p\in U$ let $F(p)$ be the set of all open $V$ such that $p\in V\in U.$ Let $G(p)=\cup F(p).$ Then $ p\in G(p) \subset U.$ And   $G(p)$ is open because it is the union of the family $F(p)$ of open sets.
(2). Therefore $$U=\cup_{p\in U}\{p\}\subset \cup_{p\in U}G(p)\subset \cup_{p\in U}U=U.$$ And $\cup_{p\in U}G(p$) is open because it is the union of the family $\{G(p):p\in G\}$ of open sets.
Applying this to your Q :For $(x,y)\in \mathbb R^2$ and any real $a,b$ the set $(-a+x,a+x)\times(-b+y,b+y)$ is open.
If $Ax+By<C$ let $C-(Ax+By)=d.$ Take $a>0$ and $b>0$ such that $|Aa|<d/2$ and $|Bb|<d/2.$ Then $(x,y)\in V=(-a+x,a+x)\times (-b+y,b+y),$ and for all $(x',y')\in V$ we have $$Ax'+By'\leq Ax+By+|A(x'-x)|+|B(y'-y)|<Ax+By+d/2+d/2=C.$$