> [!theoremb] Heine-Borel Theorem > > The following definitions of [[Compactness|compactness]] over the [[Standard Topology|standard topology]] are equivalent: > - Every [[Open Cover|open cover]] has a finite subcover. > - The set is [[Closed Set|closed]] and bounded. > - Every [[Sequence|sequence]] of points in the set has a subsequence [[Limit|converging]] to a point in it (sequential). > > *Proof*. > > **A compact set is bounded:** If a set is *not* bounded, then it is easy to generate a sequence with no converging subsequences, and create an open cover that has no finite subcover from it. Therefore a compact set must be bounded. > > **A compact set is closed:** Let $S$ be a set where all open covers of $S$ have a finite subcover. Suppose that $S^\prime$ is not [[Open Set|open]], then > $ > \exists x \in S^\prime: \forall \varepsilon > 0, B_{\varepsilon}(x) \not\subseteq S^\prime \Rightarrow \exists y \in S: d(x, y) < \varepsilon > $ > We consider the "ring" at $\varepsilon$ to be > $ > R(\varepsilon) = \bracs{\frac{\varepsilon}{4} < \bracs{z: d(x, z) < \frac{\varepsilon}{2}}: d(x, y) = \frac{\varepsilon}{2}} > $ > Then > $ > \bigcup_{k = 1}^{\infty}R\paren{\frac{1}{k}} > $ > is an open cover of $S$, since infinitely many "rings" would be required, each contributing to the union, this open cover has no finite subcovers. Therefore $S^\prime$ must be open and $S$ must be closed. A similar argument for sequences can also be constructed. > **A closed and bounded set is sequentially compact:** If a set is closed and bounded, then the convergent subsequences of any sequence would converge in the set ([[Bolzano-Weierstrass Property|Bolzano-Weierstrass Theorem]] over [[Axiom of Induction|Induction]], [[Order Limit Theorem]]). > > **A sequentially compact set is closed and bounded:** both can be done with an easy contradiction. > > **Closed intervals/boxes are compact:** First let $[a, b]$ be a closed interval, $\Sigma$ be an open cover of $[a, b]$, and let > $ > Y = \bracs{x \in [a, b]: \Sigma \text{ has a finite subcover of } [a, x]} > $ > First, $Y \ne \emptyset$ as any open set that contains $a$ is an open cover of $[a, a]$, therefore $a \in Y$. Then > $ > \begin{align*} > &\exists \sup Y \in [a, b], \exists U \in \Sigma: \sup Y \in U \\ > &\exists\varepsilon > 0: (\sup Y - \varepsilon, \sup Y + \varepsilon) \subseteq U > \end{align*} > $ > and by the property of [[Supremum and Infimum|supremum]], > $ > \exists x \in Y: x \in (\sup Y - \varepsilon, \sup Y + \varepsilon) > $ > as $\Sigma$ has a finite subcover of $[a, x]$, including $U$ into that finite subcover makes it cover $[a, \sup Y]$, therefore $\sup Y \in Y$. > > Suppose that $\sup Y < b$, then $\Sigma$ has a finite subcover of $[a, \sup Y]$. Let $U \in \Sigma: \sup Y \in U$, then > $ > \exists \varepsilon > 0: (\sup Y - \varepsilon, \sup Y + \varepsilon) \subseteq U > $ > but there would be a new finite subcover of $[a, \sup Y + \varepsilon/2]$, meaning that $\sup Y + \varepsilon/2 \in Y$, contradiction. Therefore $\sup Y = b$. > > Induction over dimensions to prove this for $\real^n$. > > **A closed subset of a compact set is compact**: Let $S$ be a compact set, $\Sigma$ be an open cover, and $T$ be a closed subset of $S$. Let $\sigma \subseteq \Sigma$ be an open cover of $T$, then $\sigma \cup \bracs{T^\prime}$ would be an open cover of $S$ ($T$ is closed). Since $\sigma \cup \bracs{T^\prime}$ must have a finite subcover of $S$, and that finite subcover must also cover $T$ ($T \subset S$), it would have a finite subcover of $T$. > > **A closed and bounded set is compact:** Let $S$ be a closed and bounded set, then it is a closed subset of some box $[-M, M]^n$, which makes it compact as well.