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\{{d \left({x,y}\right), d \left({x,z}\right)\right\}}
\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
- $d(x,w) \le d(x,y) + d(y,w) < d(x,y) + q$
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2.3$: Open sets in metric spaces