# Does totally bounded imply compact?

## Does totally bounded imply compact?

Every complete totally bounded metric space is compact. Proof Let (X, d) be a complete totally bounded metric space.

## Is a sequentially compact?

In mathematics, a topological space X is sequentially compact if every sequence of points in X has a convergent subsequence converging to a point in X.

**What is sequentially compact metric space?**

Sequentially compact metric spaces. Definition 1.1. Say a metric space X is sequentially compact if every sequence in X has a subsequence that converges in X. Lemma 1.2 (Lebesgue number lemma). Suppose K is a subset of a metric space X and K is sequentially compact.

### How do you prove a set is sequentially compact?

A metric space (X, d) is called sequentially compact if every sequence in X has a convergent subse- quence. Similarly, a subset S ⊆ X is sequentially compact if every sequence of points in S has a subsequence that converges to a point in S. By convention, the empty set ∅ is considered sequentially compact.

### How do you prove a metric space is totally bounded?

A subset A of a metric space is called totally bounded if, for every r > 0, A can be covered by finitely many open balls of radius r. For example, a bounded subset of the real line is totally bounded. On the other hand, if ρ is the discrete metric on an infinite set X, then X is bounded but not totally bounded.

**How do you prove a metric space is compact?**

Uα = X. Proposition 2.1 A metric space X is compact if and only if every collection F of closed sets in X with the finite intersection property has a nonempty intersection. points in X has a convergent subsequence.

#### Is sequentially compact the same as compact?

Theorem: A subset of a metric space is compact if and only if it is sequentially compact. Proof: If X is not sequentially compact, there exists a sequence (xn) in X that has no con- vergent subsequence. Since there is no convergent subsequence, (xn) must contain an infinite number of distinct points.

#### Is a convergent sequence compact?

There are many metric spaces where closed and bounded is not enough to give compactness, see for example . A useful property of compact sets in a metric space is that every sequence has a convergent subsequence. Such sets are sometimes called sequentially compact.

**How do you prove a compact metric space?**

A metric space X is compact if every open cover of X has a finite subcover. 2. A metric space X is sequentially compact if every sequence of points in X has a convergent subsequence converging to a point in X. 10.3 Examples.

## Are the reals compact?

The set ℝ of all real numbers is not compact as there is a cover of open intervals that does not have a finite subcover. For example, intervals (n−1, n+1) , where n takes all integer values in Z, cover ℝ but there is no finite subcover. In fact, every compact metric space is a continuous image of the Cantor set.

## Can a metric space be bounded?

A subset S of a metric space (M, d) is bounded if there exists r > 0 such that for all s and t in S, we have d(s, t) < r. A metric space is compact if and only if it is complete and totally bounded. A subset of Euclidean space Rn is compact if and only if it is closed and bounded.

**Is a bounded metric space compact?**

Metric spaces (X, d) is compact. (X, d) is complete and totally bounded (this is also equivalent to compactness for uniform spaces). (X, d) is sequentially compact; that is, every sequence in X has a convergent subsequence whose limit is in X (this is also equivalent to compactness for first-countable uniform spaces).

### How to show that a totally bounded metric space X is compact?

Show that a totally bounded complete metric space X is compact. I can use the fact that sequentially compact ⇔ compact. Attempt: Complete ⟹ every Cauchy sequence converges. Totally bounded ⟹ ∀ ϵ > 0, X can be covered by a finite number of balls of radius ϵ.

### How to show that X is totally bounded?

You need to show that if X is totally bounded, every sequence in X has a Cauchy subsequence. Let σ = ⟨ x n: n ∈ N ⟩ be a sequence in X.

**How to do a totally bounded Cauchy analysis?**

Attempt: Complete ⟹ every Cauchy sequence converges. Totally bounded ⟹ ∀ ϵ > 0, X can be covered by a finite number of balls of radius ϵ. I’m trying to show that all sequences in X have a subsequence that converges to an element in X. I don’t see how to go from convergent Cauchy sequences and totally bounded to subsequence convergent i n X.