---- > [!definition] Definition. ([[totally bounded]]) > A [[metric space]] $(X, d)$ is said to be **totally bounded** if for every $\varepsilon>0$ there is a finite [[cover|covering]] of $X$ by $\varepsilon$-[[metric topology|balls]]. > ^definition > [!note] Remarks. > - Clearly [[compact]] $\implies$ totally bounded. The [[for metric spaces, compact iff complete and totally bounded|converse]] holds if $X$ is [[complete]] ([[for metric spaces, compact iff complete and totally bounded|see]]). > - Clearly totally bounded $\implies$ [[bounded set|bounded]]. The converse holds e.g. for subsets of finite-dimensional [[norm|normed spaces]].[^1] > >- If $\text{dim }X=\infty$, e.g. the closed unit ball is not totally bounded (see [[Riesz's Lemma]]) > [!equivalence] For normed spaces. > > A subset $A$ of a [[norm|normed space]] $X$ is [[totally bounded]] if and only if $A$ is bounded and for all $\varepsilon>0$, there exists a *finite-dimensional* [[linear subspace]] $Y \leq X$ [[distance from point to set|such that]] $x_{0} \in A \implies \operatorname{dist}(x_{0},Y)<\varepsilon.$ > "Totally bounded (and in turn compactness) might not be finite dimensional, but is not too far away." > [!proof]- Proof. > Let $X$ be a normed space. > > $\implies.$ Suppose $A \subset X$ is totally bounded; let $F$, $\# F<\infty$, be an $\varepsilon$-net, $A \subset \bigcup_{x \in F}B^{\circ}_{\varepsilon}(x)$. Now $Y:=\span F$ is finite-dimensional and any $x_{0} \in A$ satisfies $\|x-y\|<\varepsilon$ for some $y \in Y$, thus the result. > > $\impliedby$. Fix $\varepsilon>0$. Fix $Y \leq X$ with $\operatorname{dist}(a, Y)<\varepsilon / 2$ for all $a \in A$. Since $A$ is bounded, have $A \subset B_{r}(0)$ for some $r>0$. Now $Y \cap B_{r}(0)$ is a closed bounded subset of a finite dimensional space, hence is [[compact]] by [[Heine-Borel theorem|Heine-Borel]] and so admits a finite $\frac{\varepsilon}{2}$-net. Now triangle. > [!basicexample] > ^basic-example Let $X$ be $\ell^{1 \leq p < \infty}$ or $c_{0}$. Translating the equivalence above, have that $A \subset X$ is totally bounded iff $A$ bounded and tails in $A$ go uniformly to zero: $\sup_{a \in A} \|a - P_{n}a\|_{p} \xrightarrow{n\to \infty}0$. [^1]: Indeed, [[norm|all norms on finite-dimensional vector spaces are topologically equivalent]] so WLOG consider $(\mathbb{R}^{n}, \|\cdot\|_{\infty})$ and a bounded subset $S$. Using boundedness, fix a cube $Q=[-R,R]^{n} \supset S$. Cleary $Q$ can covered by finitely many $\varepsilon$-cubes, thus so can be $S$. ---- #### ---- #### 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 > ```