User:Dfeuer/Open Set may not be Open Ball

[{WIP|not there yet}}

Theorem
Let $M = \left({A, d}\right)$ be a metric space with at least $3$ distinct points.

Then there exists $U \subseteq A$, where $U$ is open in $M$, but is not an open ball.

Proof
Let $x, y, z \in A$ be $3$ distinct points in $M$ such that
 * $d \left({x, y}\right) \le d \left({y,z}\right)$
 * $d \left({x, z}\right) \le d \left({y,z}\right)$

Let $r = \dfrac {\min\left\ \min\left\{{2, $.

Let $U = B_r(y) \cup B_r(z)$.

Then because the union of open sets is open, $U$ is open.

Suppose for the sake of contradiction that $U$ is an open ball.

Then there must be a point $w$ and a positive real number $q$ such that $U = B_q(w)$.

Since $w \in U$, we must have $w \in B_r(y)$ or $w \in B_r(z)$.

Suppose without loss of generality that $w \in B_r(y)$.

By the triangle inequality:
 * $d(x,w) \le d(x,y) + d(y,w) < d(x,y) + r$

We also have $d(y,z) \le d(y,w) + d(w,z) \le r + d(w,z) < r + q$.

Thus $d(w,y) < q$ and $d(w,z) < q$.

By the triangle inequality:
 * $d(y,z) < 2q$

Then by the triangle inequality:
 * $d(x,w) \le d(x,y) + d(y,w) < d(x,y) + r$.

By the definition of $r$:
 * $d(x,w) \le d(x,y) + d(y,w) < d(x,y) (1+1/4)$.

Because $r < d \left({x,y}\right) \le d \left({x,z}\right)$:
 * $x \notin B_r(y)$
 * $x \notin B_r(z)$

Thus $x \notin U$.

We will show that $x \in U$, a contradiction.

By the triangle inequality:


 * $d(y,z) \le


 * $d(x,w) \le d(x,y) + d(y,w) < d(x,y) + q$

{{qed}}