nLab
sequentially compact topological space

Context

Topology

topology (point-set topology)

see also algebraic topology, functional analysis and homotopy theory

Introduction

Basic concepts

Universal constructions

Extra stuff, structure, properties

Examples

Basic statements

Theorems

Basic homotopy theory

Contents

Idea

A topological space is called sequentially compact if every sequence of points in that space has a sub-sequence which converges. In general this concept neither implies nor is implied by that of actual compactness, but for some types of topological spaces, such as metric spaces, it is equivalent.

Compactness is an extremely useful concept in topology. The basic idea is that a topological space is compact if it isn’t “fuzzy around the edges”.

Whilst one can study a topological space by itself, it is often useful to probe it with known spaces. A common choice for topological spaces, and in particular metric spaces, is to use the natural numbers, and the 1-point compactification of the natural numbers. This is more traditionally known as studying the topology using sequences and convergent sequences.

Thus one can ask, “Can I detect compactness using probes from \mathbb{N}, and {*}\mathbb{N} \cup \{*\}?”. The short answer to this is “No”, but that just reveals that the question was too restrictive. Rather, one should ask “What does compactness look like if all I’m allowed to use are probes from \mathbb{N} and {*}\mathbb{N} \cup \{*\}?”. The answer to that question is “sequential compactness”.

Thus sequential compactness is what compactness looks like if all one has to test it are sequences.

Definition

Definition

A topological space is sequentially compact if every sequence in it has a convergent subsequence.

Properties

The following is a list of properties of and pertaining to sequentially compact spaces.

  1. For a metric space, the notions of sequential compactness and compactness coincide. See at sequentially compact metric spaces are equivalently compact metric spaces.

  2. The Eberlein–Šmulian theorem? states that in a Banach space, for a subset with regard to the weak topology, compactness and sequentially compactness are both equivalent to the weaker notion of countable compactness.

  3. A countable product of sequentially compact spaces is again sequentially compact.

    Let {X k}\{X_k\} be a countable family of sequentially compact spaces. Let (a l)(a_l) be a sequence in X k\prod X_k. For each mm we recursively define an infinite subset A mA m1A_m \subseteq A_{m-1} \subseteq \mathbb{N} with the property that the sequence (a l) lA m(a_l)_{l \in A_m} converges when projected down to k=1 mX k\prod_{k=1}^m X_k. Let l m=min{A l}l_m = \min\{A_l\}. Consider the sequence (a l m)(a_{l_m}). For each kk, we choose a limit x kx_k of the projection of (a l) lA k(a_l)_{l \in A_k} to X kX_k. Let x=(x k)X kx = (x_k) \in \prod X_k. Let UU be a neighbourhood of xx. Then there is some nn \in \mathbb{N} and neighbourhood U n k=1 nX kU_n \subseteq \prod_{k=1}^n X_k of (x k) k=1 n(x_k)_{k=1}^n such that UU contains the preimage of U nU_n. For mnm \ge n, the sequence (l m)(l_m) is contained in A nA_n and so the image of (a l m)(a_{l_m}) converges to (x k) k=1 n(x_k)_{k=1}^n. Hence there is some rr such that for mrm \ge r, the projection of a l ma_{l_m} lies in U nU_n. Hence for mrm \ge r, a l mUa_{l_m} \in U. Thus (a l m)(a_{l_m}) converges to (x k)(x_k) and so X k\prod X_k is sequentially compact.

    This shows that the example of a compact space that is not sequentially compact is about as simple as can be.

  4. The theorem that a continuous bijection from a compact space to a Hausdorff space is a homeomorphism has a counterpart for sequentially compact spaces.

    Theorem

    Let 𝒯 1\mathcal{T}_1 and 𝒯 2\mathcal{T}_2 be two topologies on a set XX such that:

    1. 𝒯 1𝒯 2\mathcal{T}_1 \supseteq \mathcal{T}_2 (equivalently, the identity map on XX is continuous as a map (X,𝒯 1)(X,𝒯 2)(X,\mathcal{T}_1) \to (X, \mathcal{T}_2))
    2. 𝒯 1\mathcal{T}_1 is sequentially compact
    3. 𝒯 2\mathcal{T}_2 is completely regular and singleton sets are G δG_\delta-sets,

    then 𝒯 1=𝒯 2\mathcal{T}_1 = \mathcal{T}_2.

    Proof

    Let VXV \subseteq X be such that V𝒯 2V \notin \mathcal{T}_2. Then it must be non-empty and there must be a point vVv \in V such that VV is not a neighbourhood of vv. As 𝒯 2\mathcal{T}_2 is completely regular and singleton sets are G δG_\delta sets, there is a continuous function g:(X,𝒯 2)g \colon (X, \mathcal{T}_2) \to \mathcal{R} such that g 1(0)={v}g^{-1}(0) = \{v\}. Since VV is not a neighbourhood of vv, for each nn \in \mathbb{N}, the set g 1(1n,1n)g^{-1}(-\frac1n, \frac1n) is not wholly contained in VV. Thus for each nn there is a point x nXx_n \in X such that x nVx_n \notin V and |g(x n)|<1n|g(x_n)| \lt \frac1n. As 𝒯 1\mathcal{T}_1 is sequentially compact, this sequence has a 𝒯 1\mathcal{T}_1-convergent subsequence, say (x n k)(x_{n_k}) converging to yy. Since g(x n)0g(x_n) \to 0, g(x n k)0g(x_{n_k}) \to 0 and thus g(y)=0g(y) = 0. Thus y=vy = v and so (x n k)v(x_{n_k}) \to v in 𝒯 1\mathcal{T}_1. As x n kVx_{n_k} \notin V for all n kn_k, and vVv \in V, it must be the case that VV is not a 𝒯 1\mathcal{T}_1-neighbourhood of vv. Hence V𝒯 1V \notin \mathcal{T}_1. Thus 𝒯 1𝒯 2\mathcal{T}_1 \subseteq \mathcal{T}_2, whence they are equal.

  5. The image of a sequentially compact space XX under a continuous map f:XYf: X \to Y is also sequentially compact. For suppose y ny_n is a sequence in f(X)f(X), say y n=f(x n)y_n = f(x_n). Then x nx_n has a convergent subsequence x n jx_{n_j}, converging to xx say, and by continuity y n j=f(x n j)y_{n_j} = f(x_{n_j}) converges to f(x)f(x).

Relationship to Compactness

Compactness does not imply sequentially compactness, nor does sequentially compactness imply compactness, without further assumptions, see at Examples and counter-examples below.

In metric spaces for example both notions coincide, see at sequentially compact metric spaces are equivalently compact metric spaces. (This is a consequence of the Lebesgue number lemma and the fact that sequentially compact metric spaces are totally bounded.)

This is not a contradiction to the statement that compact is equivalent to every net having a convergent subnet: Given a sequence in a compact space, its convergent subnet need not be a subsequence (see net for a definition of subnet).

Examples and counter-examples

A metric space is sequentially compact precisely if it is compact. See at sequentially compact metric spaces are equivalently compact metric spaces.

In general neither of these two properties implies the other:

Example

(a compact space which is not sequentially compact)

Consider the product topological space (with its Tychonoff topology)

X[0,1)Disc({0,1}) X \coloneqq \underset{[0,1)}{\prod} Disc(\{0,1\})

of copies of the discrete space on two elements, indexed by the points in the half-open interval. Since Disc({0,1})Disc(\{0,1\}) is a finite discrete topological space it is clearly compact. Therefore the Tychonoff theorem says that also XX is compact.

But here is an instance of a sequence x ()x_{(-)} in XX which does not have a convergent sub-sequence:

By the nature of the product, an element xXx \in X is a tuple of elements π r(x){0,1}\pi_r(x) \in \{0,1\} for r[0,1)r \in [0,1). Now for nn \in \mathbb{N} define x nx_n by

π r(x n)nth digit in the binary expansion ofr. \pi_r(x_n) \coloneqq n\text{th digit in the binary expansion of}\,r \,.

Suppose this sequence (x n) n(x_n)_{n \in \mathbb{N}} had a subsequence (x (n k)) k(x_{(n_k)})_{k \in \mathbb{N}}, converging to some x Xx_\infty \in X, hence that for every open neighbourhood {x }UX\{x_\infty\} \subset U \subset X there were a k 0k_0 \in \mathbb{N} such that x n kk 0Ux_{n_{k \geq k_0}} \in U.

Considering then for each r[0,1)r \in [0,1) the Tychonoff-open subset

{π r(x )}×(r[0,1)\{r}{0,1}) \{ \pi_r(x_\infty) \} \times \left( \underset{r' \in [0,1) \backslash \{r\}}{\prod} \{0,1\} \right)

this would imply that there were n 0n_0 such that

π r(x k nn 0)=π r(x ) \pi_r( x_{k_{n \geq n_0}} ) = \pi_r(x_\infty)

hence that the binary expansion of rr had the same digit in position k nk_n for all nn 0n \geq n_0. This is clearly not the case for all rr (if it is true for some rr, just change one of these digits to obtain an rr' for which it is not), and hence we have a proof by contradiction.

References

  • Buskes, van Rooij, Topological Spaces: From Distance to Neighbourhood, Springer 1997

  • Lynn Steen, J. Arthur Seebach, Counterexamples in Topology, Springer-Verlag, New York (1970) 2nd edition, (1978), Reprinted by Dover Publications, New York, 1995

  • Wayne Patty, Foundations of Topology, Jones and Bartlett Publishers (2008)

  • Stijn Vermeeren, Sequences and nets in topology, 2010 (pdf)

See also

Revised on May 15, 2017 09:09:43 by Todd Trimble (67.81.95.215)