Basics – Notes

Compactness

theorem (Folland 0.25):

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.

proposition (Compact if and only if sequentially compact for metric spaces):

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

#todo Proof

proposition (A unit ball that is not compact):

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

proof (?):

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

theorem (Heine-Borel):

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

proposition (Geometric Series):

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

corollary (?):

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

proposition (?):

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

lemma (?):

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

proposition (Finite unions of nowhere dense sets are still nowhere dense):

A finite union of nowhere dense is again nowhere dense.

theorem (Baire):

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

proposition (?):

The Cantor set is closed with empty interior.

proof (?):

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.

corollary (?):

The Cantor set is nowhere dense.

proposition (Existence of Smooth Compactly Supported 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: $\begin{align*} X \subseteq {\mathbf{R}}^n \text{ is compact } \iff X \text{ is closed and bounded} .\end{align*}$
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: ${\mathbf{R}}$ 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: ${\mathbf{R}}$ being complete is equivalent to “absolutely convergent implies convergent” for sums of real numbers.
Compacts subsets $K \subseteq {\mathbf{R}}^n$ 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