Prove the open area between two parallel lines(distinct) is an element of the usual topology on $\mathbb{R}^2$.
First I want to prove for the case if the two parallel lines are of the form $x+y = a$ and $ x+y = b, a < b$.
I need to show for each point $p$ in the area, $\exists$ $\epsilon > 0 $ for which $B_{\epsilon}(p)$ ($\epsilon$-ball around $p$) is contained in the region $x+y > a$ and $x+y < b $. I am unable to state $\epsilon$ precisely.
For each $\epsilon>0$, the $\epsilon$-ball about a point $p$ in $\mathbb{R}^2$ is $B_{\epsilon}(p) = \{y\in \mathbb{R}^2 : ||p-y||<\epsilon\}$.
How could I prove this rigorously?
The $\epsilon$ should be smaller than $\min\{d(p,\ell_1),d(p,\ell_2)\}$ where $d(p,\ell_i)$ is the distance between the point $p$ and the two respective lines $\ell_i$, $i=1,2$. There is a formula that gives this distance (see link), and in this specific question it gives $$ d(p,\ell_1) = \frac{|p_1 + p_2 + a|}{\sqrt{2}} \text{ and } d(p,\ell_2) = \frac{|p_1 + p_2 + b|}{\sqrt{2}},$$
with $p=(p_1,p_2)$.