Problem Prove that disc {$x,y: x^2 +y^2 < 1$} is an open subset of $\mathbb{R}^2$, and then that every open disc in the plane is an open set.
Steps involved
(i) Let $a,b$ be any point in the disc
$D$ = {$x,y: x^2 +y^2 < 1$}.Put $r=(x^2+y^2)^{1/2}$ .Let $R_{(a,b)}$ be the open rectangle with vertices at the points $a± \frac{1−r}{8} , b± \frac{1−r}{8}$. Verify that $R_(a,b)⊂ D$.
(ii) Using (i) show that $D= \cup R_{(a,b)}$. $a,b ∈D$
(iii) Deduce from (ii) that D is an open set in $\mathbb{R}^2$.
(iv) Show that every disc
{$(x,y): (x−a)^2 +(y −b)^2 < c^2, a,b,c ∈ \mathbb{R}$}
Attempt
To prove step 1, i am trying to show that $(a+\frac{1-r}{8})^2 + (b+\frac{1-r}{8})^2 \leq 1$ but i couldn't prove it .
$(a+\frac{1-r}{8})^2 + (b+\frac{1-r}{8})^2 \leq (a+\frac{1}{8})^2 + (b+\frac{1}{8})^2 .$. How to proceed.
NB - I am more comfortable using neighborhood around (x,y) and proving it.
The topology on the plane is given by the metric $d$ presumably ($d$ is the Euclidean square root metric, $d((x_1,x_2),(y_1,y_2)) = \sqrt{((x_1 - y_1)^2 + (x_2-y_2)^2}$).
The open unit ball $B$ is just $B_d((0,0),1)$, the $d$-ball around the origin $0$ with radius $1$ (as $x^2 + y^2 < 1$ iff $d((x,y),(0,0)) < 1$). And in a metric space any open ball is open in the metric topology:
Let $(x,y) \in B$. Then $d((x,y), (0,0)) < 1$ so define $r = 1 - d((x,y), (0,0)) > 0$. Then $B_d((x,y), r) \subseteq B$ because if $(u,v) \in B_d((x,y), r)$ then by the triangle inequality:
$$d((u,v),(0,0)) \le d((u,v), (x,y)) + d((x,y),(0,0)) < r + d((x,y), (0,0)) = 1$$
so that $(u,v) \in B$.
This argument works in any metric space: all open balls are indeed open (i.e. in the metric topology), hence the name "open" ball, of course.