# 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. 24/04/2021