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 {1x} \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 welldefined 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

BolzanoWeierstrass: Every bounded sequence has a convergent subsequence.

HeineBorel: \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 nowheredense 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