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)

Symbols:LaTeX Commands/ProofWiki Specific

Theorem
Let $\openint a b$ be an open real interval.

Let $c \in \openint a b$.

Let $f: \openint a b \setminus \set c \to \R$ be a real function.

Let $L \in \R$.

Suppose that:


 * $\forall n > 0 \in \N: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < 2^{-n} $

Then the limit of $f$ exists as $x$ tends to $c$, and is equal to $L$.

Proof
Let the limit of $f$ as $x \to c$ exist and equal $L$, as described in the definition of limit:


 * $\forall \epsilon > 0 \in \R: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$.

Denote by $\map P {\epsilon,\delta}$ the proposition that the above statement holds.

From Limit with Rational Epsilon and Delta, it is sufficient to only consider $\epsilon \in \Q$.

We have that the Rational Numbers are Countably Infinite.

Also, the Unit Interval is Uncountable.

Thus $\openint 0 1 \cap \Q$ is countably infinite. index them by $i \in \N$.

Impose a well-ordering on $\openint 0 1 \cap \Q$; This is possible from the Well-Ordering Theorem.

Then we can recursively define a sequence of $\sequence {2_n}$ such that:

For every $\epsilon \in \openint 0 1 \cap \Q$, we map said $\epsilon$ to powers of $2$ as follows:

For $\epsilon_i \in \openint 0 1 \cap \Q \setminus \set{\epsilon_{i-1},\epsilon_{i-2}, \ldots}$ is mapped to $2^{-n_i}$ for some $n$ Definition:Sufficiently Large satisfying $2^{-n_i} < \epsilon_i$


 * and:

$2^{-n_i} < 2^{-n_j}$ for all $i > j$.

Such a sequence is a subsequence of $\sequence{2^{-n}}_{n \in \N}$

bleh... this should be a lot shorter and simpler, I feel like I'm missing some obvious shortcut. Most places I've seen take this for granted. I work better during mornings, I'll look again tomorrow. Feel free to leave feedback. I'm overthinking something. --GFauxPas (talk) 19:34, 8 February 2020 (EST)


 * Also, I can't find a page that examines the convergence of the sequence $\sequence{p^{-n}}_{n \in \N}$ for $|p|<1, |p|>1$. --GFauxPas (talk) 19:39, 8 February 2020 (EST)

Theorem
Let $\map \BB {\R, \size {\, \cdot \,} }$ be the Borel Sigma-Algebra on $\R$ with the Usual Topology.

Let $f: \R \to \R$ be a Continuous Real Function.

Let $\map D f$ be the set of all points at which $f$ is differentiable.

Then $\map D f$ is a Borel Set with respect to $\map \BB {\R, \size {\, \cdot \,} }$.

Proof
Let $F: \R \times \R \setminus \set 0 \to \R$ be defined as:


 * $\map F {x, h} = \dfrac {\map f {x + h} - \map f x} h$

By the definition of derivative:


 * $\map {f'} x$ exists




 * $\displaystyle \lim_{h \mathop \to 0} \map F {x, h}$ exists.

The second limit can written as:


 * $\exists y \in \R: \forall \varepsilon > 0: \exists \delta > 0: \forall h \in \R \setminus \set{0}: \size h < \delta \implies \size {\map F {x, h} - y} < \epsilon$

Let $\sequence {y_n}: \Q \to \R$ be a Rational Sequence converging to the $y$ whose existence is asserted such.

Such a sequence exists by Rationals are Everywhere Dense in Reals.

[[Category:Continuity [[Category:Sigma-Algebras