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By definition, if $S$ is totally bounded it is covered by a finite number of open balls radius $\epsilon/2$ centered on points in $S$. Closed balls of the same radius $\epsilon/2$ on the same centres therefore also cover $S$. Their finite union, $C$ is a closed set containing $S$.
If a subset $A \subset X$ is totally bounded, then its closure $\overline{A}$ is totally bounded. Definition of "totally bounded": A set $A$ is totally bounded if, for each $\varepsilon > 0$ , there is a finite $F\subset A$ such that $A \subset \bigcup_\limits{x \in F} B(x, \varepsilon) $ .
You need to show that if $X$ is totally bounded, every sequence in $X$ has a Cauchy subsequence. Let $\sigma=\langle x_n:n\in\mathbb{N}\rangle$ be a sequence in $X$.
A metric space is said to be totally bounded if every sequence admits a Cauchy subsequence; in complete metric spaces, a set is compact if and only if it is closed and totally bounded. Each totally bounded space is bounded (as the union of finitely many bounded sets is bounded).
21 de dic. de 2020 · Definitions Bounded set: Let <M,ρ> be a metric space. We say that the subset A of M is bounded if there exists a positive number L such that ρ(x,y)≤L (x,y∈A) (Wikipedia)Diameter of a set: If A is bounded, we define diameter of A (denoted by diam A) as diam A=$latex {l.u.b.}_(x,y\in A)$ ρ(x,y).
A closed subset of a complete metric space is itself complete, when considered as a subspace using the same metric, and conversely. Note that this means, for example, that a closed interval in R is a complete metric space. Theorem 5.3: Let ( M, d) be a complete metric space, and let X be a subset of M.
12 de dic. de 2013 · The metric totally-bounded spaces, considered as topological spaces, exhaust all regular spaces (cf. Regular space) with a countable base. A subspace of a Euclidean space is totally bounded if and only if it is bounded.
