# Basics

## Compactness

For $$E \subseteq (X, d)$$ a metric space, TFAE:

• $$E$$ is complete and totally bounded.
• $$E$$ is sequentially compact: Every sequence in $$E$$ has a subsequence that converges to a point in $$E$$.
• $$E$$ is compact: every open cover has a finite subcover.

Note that $$E$$ is complete as a metric space with the induced metric iff $$E$$ is closed in $$X$$, and $$E$$ is bounded iff it is totally bounded.

## Topology / Sets

Metric spaces are compact iff they are sequentially compact, (i.e. every sequence has a convergent subsequence).

#todo Proof

The unit ball in $$C([0, 1])$$ with the sup norm is not compact.

Take $$f_k(x) = x^n$$, which converges to $$\chi(x=1)$$. The limit is not continuous, so no subsequence can converge.

$$X\subseteq {\mathbb{R}}^n$$ is compact $$\iff X$$ is closed and bounded.

\begin{align*} \sum_{k=0}^\infty x^k = \frac 1 {1-x} \iff {\left\lvert {x} \right\rvert} < 1 .\end{align*}

\begin{align*} \sum_{k=0}^\infty \frac 1 {2^k} = 1 .\end{align*}

Singleton sets in $${\mathbb{R}}$$ are closed, and thus $${\mathbb{Q}}$$ is an $$F_\sigma$$ set.

Any nonempty set which is bounded from above (resp. below) has a well-defined supremum (resp. infimum).

## Smallness for sets

A finite union of nowhere dense is again nowhere dense.

$${\mathbb{R}}$$ is a Baire space, i.e. $${\mathbb{R}}$$ can not be written as a countable union of nowhere dense sets.

The Cantor set is closed with empty interior.

Its complement is a union of open intervals, and can’t contain an interval since intervals have positive measure and $$m(C_n)$$ tends to zero.

The Cantor set is nowhere dense.

## Smallness for functions

There exist smooth compactly supported functions, e.g. take \begin{align*} f(x) = e^{-\frac{1}{x^2}} \chi_{(0, \infty)}(x) .\end{align*}

• Arzela - Ascoli 1: If $$\mathcal{F}$$ is pointwise bounded and equicontinuous, then $$\mathcal{F}$$ is totally bounded in the uniform metric and its closure $$\overline{\mathcal{F}} \in C(X)$$ in the space of continuous functions is compact.

• Arzela - Ascoli 2: If $$\left\{{f_k}\right\}$$ is pointwise bounded and equicontinuous, then there exists a continuous $$f$$ such that $$f_k \xrightarrow{u} f$$ on every compact set.

#todo Proof

• Bolzano-Weierstrass: Every bounded sequence has a convergent subsequence.

• Heine-Borel: \begin{align*} X \subseteq {\mathbb{R}}^n \text{ is compact } \iff X \text{ is closed and bounded} .\end{align*}

• Baire Category Theorem: If $$X$$ is a complete metric space, then $$X$$ is a Baire space:

• For any sequence $$\left\{{U_k}\right\}$$ of open, dense sets, $$\cap_k U_k$$ is also dense.
• $$X$$ is not a countable union of nowhere-dense sets
• Nested Interval Characterization of Completeness: $${\mathbb{R}}$$ being complete $$\implies$$ for any sequence of intervals $$\left\{{I_n}\right\}$$ such that $$I_{n+1} \subseteq I_n$$, $$\cap I_n \neq \emptyset$$.

• Convergence Characterization of Completeness: $${\mathbb{R}}$$ being complete is equivalent to “absolutely convergent implies convergent” for sums of real numbers.

• Compacts subsets $$K \subseteq {\mathbb{R}}^n$$ are also sequentially compact, i.e. every sequence in $$K$$ has a convergent subsequence.

• Closed subsets of compact sets are compact.

• Every compact subset of a Hausdorff space is closed

• Urysohn’s Lemma: For any two sets $$A, B$$ in a metric space or compact Hausdorff space $$X$$, there is a function $$f:X \to I$$ such that $$f(A) = 0$$ and $$f(B) = 1$$.

• Continuous compactly supported functions are

• Bounded almost everywhere

• Uniformly bounded

• Uniformly continuous

Proof:

• Uniform convergence allows commuting sums with integrals

#todo