User:Dfeuer/Open Set may not be Open Ball

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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$ $\blacksquare$[[Category:Proven Results]] =='"`UNIQ--h-2--QINU`"' Sources == * 1975: [[Mathematician:W.A. Sutherland|W.A. Sutherland]]: [[Book:W.A. Sutherland/Introduction to Metric and Topological Spaces|''Introduction to Metric and Topological Spaces'']] ... [[Half-Open Real Interval is not Open Set|(previous)]] ... [[Empty Set is Open in Metric Space|(next)]]: $2.3$: Open sets in metric spaces