Open Set may not be Open Ball

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Theorem

Let $M = \struct {A, d}$ be a metric space.

Let $U \subseteq M$ be an open set of $M$.


Then it is not necessarily the case that $U$ is an open ball of some $x \in A$.


Proof

Consider the Euclidean space $\R^2$.

Let:

$U \subseteq \R^2: U = \set {\tuple {x_1, x_2}: a < x_1 < b, c < x_2 < d}$

Let $x = \tuple {x_1, x_2} \in U$.

Then $\map {B_\epsilon} x \subseteq U$ when $\epsilon = \min \set {x_1 - a, b - x_1, x_2 - c, d - x_2}$:

NeighborhoodInOpenSet.png

So by definition, $U$ is open in $M$.

However, $U$ is not an open ball.

$\blacksquare$


Sources