Basics – Notes

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.

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).

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.

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