> [!quote] > > Totally bounded is the idea of having a "finite amount of space". > [!definition] > > Let $(X, d)$ be a [[Metric Space|metric space]] and $E \subseteq X$. Then the following are equivalent: > - Any [[Sequence|sequence]] $\bracs{x_n}$ with $x_n \in E$ has a [[Cauchy Sequence|Cauchy subsequence]] ([[Bolzano-Weierstrass Property|Bolzano-Weierstrass style]]). > - For any $\varepsilon > 0$, $E$ can be [[Open Cover|covered]] by finitely many [[Open Set|open]] balls of radius $\varepsilon$ centred in $E$. > > If $E$ satisfies any one of the above, then $E$ is **totally bounded**. > > *Proof*. Suppose that for any $\varepsilon > 0$, $E$ can be covered by finitely many open balls of radius centred in $E$. > > First let $\varepsilon = \frac{1}{2}$, then $E$ can be covered by finitely many open balls of radius $\varepsilon$. Since $\bracs{x_n}$ is a sequence in $E$, at least one of those balls $S_1 = B_1$ must contain infinitely elements of $\bracs{x_n}$. Take the first of such element as $x_{n_1}$. > > Then let $\varepsilon = \frac{1}{2^{k}}$, then $E$ can again be covered by finitely many open balls of radius $\varepsilon$. Since $S_1$ contains infinitely elements of $\bracs{x_n}$, the intersection of one of these balls $S_k = S_{k - 1} \cap B_k$ must contain infinitely many elements of $\bracs{x_n}$. Take the first of such element whose index is greater than $n_{k - 1}$ as $x_{n_k}$. > > We have inductively constructed a subsequence $\bracs{x_{n_k}}$. Since $x_{n_{k}} \in B_k \forall k \in \nat$, we have > $ > d\paren{x_{n_j}, x_{n_k}} < \frac{1}{2^K} \quad \forall j, k \ge K > $ > which means that > $ > \forall \varepsilon >0, \exists K \in \nat, \frac{1}{2^K} < \varepsilon: d\paren{x_{n_j},x_{n_k}} < \frac{1}{2^K} < \varepsilon \quad \forall j, k \ge K > $ > and we have constructed a Cauchy subsequence. > > Now suppose that every sequence $\bracs{x_n}$ in $E$ has a Cauchy subsequence. For the sake of contradiction, suppose that there exists $\varepsilon > 0$ such that $E$ *cannot* be covered by finitely many open balls of radius $\varepsilon$. > > We again inductively construct a sequence as follows: First take any point in $E$ and label it $x_1$. Let $B_1 = B(x_1, \varepsilon)$. Now, since $E$ cannot be covered by finitely many balls of radius $\varepsilon$, > $ > E_k = E \setminus \bigcup_{j = 1}^{k}B_k \ne \emptyset > $ > We can take any element in $E_{k - 1}$ and label it $x_k$, then let $B_k = B(x_k, \varepsilon)$, and repeat the process. > > Now we have constructed a sequence $\bracs{x_{n}}$. However, for any $j, k \in \nat$, we have $x_j \not\in B(x_k, \varepsilon)$ or $d(x_j, x_k) \ge \varepsilon$. Therefore $\bracs{x_n}$ has no Cauchy subsequence. This contradicts with the fact that every sequence in $E$ has a Cauchy subsequence, which means that our assumption that $E$ cannot be covered by finitely many open balls of radius $\varepsilon$ is false. Therefore $E$ can be covered by finitely many open balls of radius $\varepsilon$.