Open Ball in Euclidean Plane is Interior of Circle

Theorem
Let $\R^2$ be the real number plane considered as a metric space under the usual metric.

Let $x = \left({x_1, x_2}\right) \in \R^2$ be a point in $\R^2$.

Let $B_\epsilon \left({x}\right)$ be the open $\epsilon$-ball at $x$.

Then $B_\epsilon \left({x}\right)$ is the interior of the circle whose center is $x$ and whose radius is $\epsilon$.

Proof
Let $S = B_\epsilon \left({x}\right)$ be an open $\epsilon$-ball at $x$.

Let $y = \left({y_1, y_2}\right) \in B_\epsilon \left({x}\right)$.

Then:

But from Equation of Circle:
 * $\left({y_1 - x_1}\right)^2 + \left({y_2 - x_2}\right)^2 = \epsilon^2$

is the equation of a [Definition:Circle|circle]] whose center is $\left({x_1, x_2}\right)$ and whose radius is $\epsilon$.

The result follows by definition of interior and Open Ball of Point Inside Open Ball.