User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

Theorem
You can cut up an $n$-cube into a bunch of little cubes

Bounded Subspace of Euclidean Space is Totally Bounded
Let $M$ be a subset of $\left({\R^n,\Vert \cdot \Vert}\right)$, where $\Vert \cdot \Vert$ is the usual metric.

If $M$ is bounded, it is totally bounded.

Proof
AS $M$ is bounded, it has a finite diameter $R$.

As $R \le 2R\sqrt{n}$, $M$ can be placed inside an $n$-cube of the form $\left[ {c-R\,.\,.\,c+R} \right]^n$.

For any $\epsilon > 0$, pick an integer $k > \sqrt{n}R \epsilon^{-1}$.

Consider a normal subdivision of $\left[ {-R\,.\,.\,R} \right]$ into $k$ pieces.

The length of each subinterval is then:


 * $\left({c+R - (c-R)}\right)k^{-1} = 2Rk^{-1}$.

By diameter of $n$-cube, each sub-cube has diameter $2Rk^{-1}\sqrt{n}$.

By the choice of $k$, $ 2Rk^{-1}\sqrt{n} < 2\epsilon$.

That means that each sub-cube can fit inside an $n$-ball of radius $\epsilon$.

This in turn implies that $M$ can be covered by a finite number of balls with radius $\epsilon$.

Thus $M$ is totally bounded by definition.

Eventually
User:GFauxPas/Sandbox/Zeta2/lnxln1-x/existence

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/integrand

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/evaluation

User:GFauxPas/Sandbox/Zeta2/FourierSeries/

User:GFauxPas/Sandbox/Zeta2/Informal Proof

User:GFauxPas/Sandbox/NumberTheory