---- > [!definition] Definition. ([[compact]]) > A [[topological space]] $X$ is said to be **(quasi-)compact** if every [[cover|open covering]] $\mathscr{A}$ of $X$ contains a finite subcollection that also covers $X$ ('finite subcover'). > [!equivalence] > $X$ is [[compact]] if and only if any cover of $X$ by [[basis for a topology|basic]] [[topology generated by a basis|open sets]] has a finite subcover. > > > [!proof]- Proof of Equivalence. > > One direction is immediate. For the other, suppose that basic open covers have finite subcovers. Let $\{ U_{\alpha} \}$ be an arbitrary open cover of $X$. Each $U_{\alpha}=\bigcup_{\beta}B_{\alpha \beta}$ for basic opens $B_{\alpha \beta}$, yielding an open cover $X=\bigcup_{\alpha} \bigcup_{\beta} B_{\alpha \beta}$ having finite subcover $X=\{ B_{i} \}_{i=1}^{n}$. Now, each $B_{i}$ belongs to some $U_{\alpha}=U_i$, $B_{i} \subset U_{i} \subset X$, and hence $X=\bigcup_{i=1}^{n}U_{i}$. > > > ^c1aa5b > [!equivalence] Equivalence for metrizable spaces. > If $X$ is [[metrizable]], then the following are equivalent: > 1. $X$ is [[compact]]; > 2. $X$ is [[sequentially compact]]; > 3. $X$ is [[limit point compact]] ([[for metrizable spaces, compact iff limit point compact iff sequentially compact|here]]) > 4. $X$ is [[complete]] and [[totally bounded]] ([[for metric spaces, compact iff complete and totally bounded]]) ^equivalence ![[compactness characterization for subspaces#^9fd510]] > [!basicexample] > - The following subspace of $\mathbb{R}$ is compact: $X=\{ 0 \} \cup \{ 1 / n : n \in \mathbb{N} \}.$ Indeed, given an [[cover|open covering]] $\mathscr{A}$ of $X$, there is an element $U \in \mathscr{A}$ containing $\{ 0 \}$. Since $\left( \frac{1}{n} \right)_{n \in \mathbb{N}} \to 0$, all but finitely many elements of $\{ 1 / n: n \in \mathbb{N} \}$ lie in $U$, and hence by taking the union of $U$ with one member of $A$ containing each of the finite remaining elements we obtain a finite subcover. >- A [[topological space]] containing finitely many points is necessarily compact— every open covering is finite (as a subset of $\mathcal{P}(X)$). >- The [[Heine-Borel theorem|Heine-Borel Theorem]] says that any closed + bounded subset of $\mathbb{R}^{n}$ with the Euclidean or sup metric is compact. > [!basicnonexample] > - $\mathbb{R}$ is not compact, for the covering of it by [[open interval]]s $\mathscr{A}:=\{ (n,n+2) : n \in \mathbb{Z}\}$ contains no finite subcover. > >- The infinite-dimensional sphere $\mathbb{S}^{\mathbb{N}}_{p}=\{ (x_{1},x_{2},\dots) \in \mathbb{R}^{\mathbb{N}}: \|(x_{1},x_{2},\dots)\|_{p}=1 \}$ is closed (it equals the preimage of $\{ 1 \}$ under the continuous map $\|\cdot\|_{p}$) and bounded but not compact in $\ell^{p}(\mathbb{R})$ for any $1 \leq p \leq \infty$. (Contrast to the finite-dimensional case.) > > > > [!proof]- Proof. > > Indeed, recall that [[for metrizable spaces, compact iff limit point compact iff sequentially compact]]. $\mathbb{S}^{\mathbb{N}}_{p}$ is not [[sequentially compact]]: if $1 \leq p < \infty$ one has $\|e_{n}\|_{p}=1 \text{ and } \|e_{n}-e_{m}\|_{p}=2^{1/p} \ (n \neq m),$ > > and if $p=\infty$ one has $\|e_{n}-e_{m}\|_{\infty}=1,$ > > and so the [[sequence]] $(e_{1},e_{2},\dots)$ has no [[converge|convergent]] [[subsequence]] (e.g. since a convergent sequence is [[Cauchy sequence|Cauchy]] and there is no Cauchy subsequence). > ^33e88d ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` # Math 395 Definition Let $X$ be a [[metric space]]. $Y \subset X$ is called *compact* if — for every collection of [[open set]]s whose union is $Y$, there is a finite subcollection whose union is *again* $Y$. ## Equivalently (the notion of 'covering') $Y \subset X$ is compact **iff** for all collections of [[open set]]s in $X$ whose union contains $Y$ there exists a finite subcollection whose union *again* contains $Y$ # More Equivalencies Let $K \subset \mathbb{R}^n$. Then $K$ is *compact* **iff** $K$ is a [[bounded set]] and a [[closed set]] in $\mathbb{R}^n$ (Heine-Borel) **iff** $K$ is [[sequentially compact]]