Compact Sets in R

Subjects: Real Analysis
Links: Open and Closed Sets in R

Def: A set $K\subseteq \mathbb{R}$ is sequentially compact if for every sequence in $K$ has a subsequence that converges to a limit that is also in $K$.

Def:**** A set $A\subseteq \mathbb{R}$ is bounded if there’s $M>0$ such that $|a|\le M$ for all $a\in A$

Bolzano-Weiestrass Theorem

Let $A\subseteq \mathbb{R}$ be a infinite bounded set, then has at least a limit point, i.e. $A^{\prime }\ne \varnothing$. This is a reformulation of the Bolzano-Weiestrass theorem for sequences.

Th (Characterization of Sequential Compactness in $\mathbb{R}$): A set $K\subseteq \mathbb{R}$ is sequentially compact iff it is closed and bounded.

Open Covers

Def: Let $A\subseteq \mathbb{R}$. An open cover for $A$ is a (possibly infinite) collection of open sets $\{U_{\lambda }\mid \lambda \in \mathrm{\Lambda }\}$ whose union contains the set $A$; that is $A\subseteq \bigcup _{\lambda \in \mathrm{\Lambda }}{U}_{\lambda }$. Given an open cover for $A$, a finite subcover is a finite subcollection of open sets from the original open cover whose union still manages to completely contain $A$.

Def: A set $K\subseteq \mathbb{R}$ is called compact if for every open cover there’s a finite subcover

Heine-Borel Theorem

Let $K\subseteq \mathbb{R}$. All of the following statements are equivalent:

• $K$ is sequentially compact
• $K$ is compact
• $K$ is closed and bounded

Nested Compact Set Property

Let $(K_n{)}_{n\in \mathbb{N}}$ be a nested sequence of nonempty compact sets, then

This is equivalent to the completeness of R

Def: A system of sets $S$ has the finite intersection property if every nonempty finite subsystem of $S$ has nonempty intersection.

Any system of compact intervals in $\mathbb{R}$ with the finite intersection property has nonempty intersection

Any system of compact sets in $\mathbb{R}$ with the finite intersection property has nonempty intersection