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)

Bounded Subspace of Euclidean Space is Totally Bounded
Let $M$ be a subspace of $\R^n$ considered as a Euclidean space with the usual metric.

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

Proof
By the definition of bounded, to say that $X$ is bounded is to say that:


 * $\exists K \in \R: \forall x, y \in M: \left \Vert{ x-y } \right \Vert \le K$.

Then,

Note $n^{1/2}$ is constant.

Write $v = x - y$ and $v_i = x_i - y_i$.

Then any ball:


 * $\left\{ { v : \Vert v \Vert < K } \right\}$

is contained in some set

$\left\{ { v : \max_i \left \vert {v_i} \right \vert < R } \right\}$

for $R$ sufficiently large.

This latter set is a hypercube, because it can be written as:


 * $[-R\,.\,.\,R]^n$

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