ProofWiki:Sandbox

Proof of Open Set may not be Open Ball
Suppose $ d(x,y) $ only can assume at most two different values for $x \ne y$.

That is, there exist $ d_1, d_2 \in ] 0,\infty [ $ with $ d_1 \le d_2 $, so $ d(x,y) = d_1 $ or $ d(x,y) = d_2 $ for all $ x , y \in M $ with $ x \ne y $.

Find two distinct points $ x, y \in M $, so $ d(x,y) = d_2 $.

Let $ U = B_{ d_1 / 2 }( x ) \cup B_{ d_1 / 2 }( y ) = \{ x, y \} $, so $ U $ is open.

Let $ a \in M $ and $ \epsilon > 0 $.

If $ a \notin M $, we have $ B_{ \epsilon }( a ) \ne U $, as $ a \in B_{ \epsilon }( a ) $.

If $ a = x $, we have $ y \notin B_{ \epsilon }( a ) $ if $ \epsilon < d_2 $, and $ B_{ \epsilon }( a ) = M $ if $ \epsilon \ge d_2 $.

In both cases, we have $ B_{ \epsilon }( a ) \ne U $.

A symmetry argument shows that if $ a = y $, we have $ B( a; \epsilon ) \ne U $.

Suppose instead that $ d(x,y) $ can assume more than two different values for $ x \ne y $.

Then we can find three distinct points $ x, y, z \in M $, so $ d(x,y) > d(x,z) $, and $ d(x,y) > d(y,z) $.

Let $ r = \min ( \frac{ d(x,y) - d(x,z) }{ 2 }, d(x,z) ) $, and $ s = \min ( \frac{ d(x,y) - d(y,z) }{ 2 } , d(y,z) ) $.

Let $ U = B_r ( x ) \cup B_s ( y ) $.

Then $ U $ is open, and $ z \notin U $.

If $ a \notin M $, we have $ B_{ \epsilon } ( a ) \ne U $, as $ a \in B_{ \epsilon ]( a ) $.

If $ a \in M $, then $ a \in B_r ( x ) $ or $ a \in B_s ( y ) $.

Suppose $ a \in B_x $, then

What I'm about to write has nothing to do with this proof; I'm just trying out some math notation. Here's an inline equation $u = \phi \cdot \exp \left\{ {{\textstyle{1 \over 2}}\sigma \left( {x + y} \right)} \right\} % MathType!MTEF!2!1!+- % feaagGart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbdfgBPj % MCPbqedmvETj2BSbqefm0B1jxALjhiov2DGCKCLv2AGW0B3bqefqvA % Tv2CG4uz3bIuV1wyUbqee0evGueE0jxyaibaieYhf9irFfeu0dXdh9 % vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqai-hGuQ8kuc9pgc9 % q8qqaq-dir-f0-yqaiVgFr0xfr-xfr-xb9adbaqaaeGaciGaaiaabe % qaamaaeaqbaaGcbaGaamyDaiabg2da9iabew9aMjabgwSixlGacwga % caGG4bGaaiiCamaacmaabaWaaSqaaSqaaiaaigdaaeaacaaIYaaaaO % Gaeq4Wdm3aaeWaaeaacaWG4bGaey4kaSIaamyEaaGaayjkaiaawMca % aaGaay5Eaiaaw2haaaaa!4F92! $ and a display equation $$\sum\nolimits_{n = 1}^N {\frac{1}} % MathType!MTEF!2!1!+- % feaagGart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbdfgBPj % MCPbqedmvETj2BSbqefm0B1jxALjhiov2DGCKCLv2AGW0B3bqefqvA % Tv2CG4uz3bIuV1wyUbqee0evGueE0jxyaibaieYhf9irFfeu0dXdh9 % vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqai-hGuQ8kuc9pgc9 % q8qqaq-dir-f0-yqaiVgFr0xfr-xfr-xb9adbaqaaeGaciGaaiaabe % qaamaaeaqbaaGcbaWaaabmaeaadaWcaaqaaiaaigdaaeaacaWGUbGa % eyyXIC9aaeWaaeaacaWGUbGaey4kaSIaaGymaaGaayjkaiaawMcaai % abgwSixpaabmaabaGaamOBaiabgUcaRiaaikdaaiaawIcacaGLPaaa % cqGHflY1daqadaqaaiaad6gacqGHRaWkcaaIZaaacaGLOaGaayzkaa % aaaaWcbaGaamOBaiabg2da9iaaigdaaeaacaWGobaaniabggHiLdaa % aa!5720! $$ Matrices can be especially problematic. Let's try some. Here's a simple one $\left[ {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right]$ and a simple matrix multiplication $$\left[ {\begin{array}{*{20}{c}}1&2&3\\4&5&6\end{array}} \right] \cdot \left[ {\begin{array}{*{20}{c}}1&1\\2&2\\3&3\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{14}&{14}\\{32}&{32}\end{array}} \right] % MathType!MTEF!2!1!+- % feaagGart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbdfgBPj % MCPbqedmvETj2BSbqefm0B1jxALjhiov2DGCKCLv2AGW0B3bqefqvA % Tv2CG4uz3bIuV1wyUbqee0evGueE0jxyaibaieYhf9irFfeu0dXdh9 % vqqj-hEeeu0xXdbba9frFj0-OqFfea0dXdd9vqai-hGuQ8kuc9pgc9 % q8qqaq-dir-f0-yqaiVgFr0xfr-xfr-xb9adbaqaaeGaciGaaiaabe % qaamaaeaqbaaGcbaWaamWaaeaafaqabeGadaaabaGaaGymaaqaaiaa % ikdaaeaacaaIZaaabaGaaGinaaqaaiaaiwdaaeaacaaI2aaaaaGaay % 5waiaaw2faaiabgwSixpaadmaabaqbaeqabmGaaaqaaiaaigdaaeaa % caaIXaaabaGaaGOmaaqaaiaaikdaaeaacaaIZaaabaGaaG4maaaaai % aawUfacaGLDbaacqGH9aqpdaWadaqaauaabeqaciaaaeaacaaIXaGa % aGinaaqaaiaaigdacaaI0aaabaGaaG4maiaaikdaaeaacaaIZaGaaG % OmaaaaaiaawUfacaGLDbaaaaa!54C7! $$

$$\begin{array}{*{20}{c}}a&b\\c&d\end{array}$$