# Prove that open half planes are open sets

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.

• Are you familiar with the concept of continuity?
– CIJ
Oct 12 '15 at 16:58
• Let $(x,y)$ be any point in a open half plane. Can you find positive $r$ small enough to ensure that the circle of radius $r$ centered at $(x,y)$ is completely contained in the open half plane? Oct 12 '15 at 16:59

A set is open if for any point in the set, you can enclose that point in an open ball that is itself contained completely in the set. The work of this problem will be to figure out an algorithm to construct an open set given any point in the half plane. So start by picking an arbitrary point $(a,b)$ in the half plane, $H = \{(x, y)\in \Bbb{R}^2\vert \space Ax + By<C\}$.

I'll work through the case of the half consisting of all $(x,y) \in \Bbb{R}^2$ where $x>0$. I believe this corresponds to having values of $A = -1, B=0,C=0$. Now if we take $(a,b) \in H$ we can use the vertical line $x = 0$ as a boundary for an open ball. We know $(a,b)$ is at minimum a distance of $a$ away from the line $x = 0$, so if we take an open ball $$B_{\frac{a}{2}}\left((a,b)\right) = \left\{(c,d) \in \mathbb{R}^2\Big\vert \space (a-c)^2+(b-d)^2<\frac{a}{2}\right\}$$ of radius $\frac{a}{2}$ centered at $(a,b)$ we are guaranteed that the open ball will be completely contained in $H$. Hence, $H$ (with the $A,B,C$ I chose) is open. Can you generalize my approach and show $H$ must always be open?

A half-plane is a translation of a set of the form

$$A_y=\{x \in \mathbb{R}^2 \mid \langle x, y \rangle\} < 0$$

for some $y \neq 0$. Note that this is equal to

$$\langle \cdot, y\rangle ^{-1}\big( \left(-\infty,0\right)\big).$$

Since $\langle \cdot, y\rangle$ is continuous and translations are homeomorphisms, the result follows.

Also, one could simply take your definition of open half plane and consider the continuous function

$$f: \mathbb{R}^2 \rightarrow \mathbb{R}$$ $$(x,y) \mapsto Ax+By-C.$$

We have that the half-plane is $f^{-1}\big( (-\infty, 0)\big)$.

• I think, this solution is contains more complicated claims than the problem itself. You know proving that the half plane is open is easier than proving inner product is continuous and translations are homeomorphisms. Oct 12 '15 at 19:39
• @muratguner Those are elementary and non-specific-for-exercises facts. MSE is intended to have different answers and approaches for a given question. Since one was already adressed by another answer, I gave this one. Also, if someone underwent linear algebra and a modicum of topology, I guarantee that this person has seen the facts that I listed (which are general, useful facts), but I don't guarantee that he has come across the specific problem given by OP. Oct 12 '15 at 19:49
• You are right. I just thought that for someone who asked this question your answer might be confusing and overwhelming. But it is always nice to have different approaches. Oct 12 '15 at 20:01

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 $$Ax'+By'\leq Ax+By+|A(x'-x)|+|B(y'-y)|<Ax+By+d/2+d/2=C.$$