nLab
sequentially compact space

Contents

Idea

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.

  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.

Relationship to Compactness

Compactness does not imply sequentially compactness, nor does sequentially compactness imply compactness, without further assumptions (see for example wikipedia: compact spaces). In metric spaces for example both notions coincide.

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).

A Compact Space that is not Sequentially Compact

A famous example of a space that is compact, but not sequentially compact, is the product space

{0,1} I:={0,1} [0,1] \{0,1\}^{I} := \{0, 1\}^{[0, 1]}

with the product topology. It is compact by the Tychonoff theorem.

Points of {0,1} I\{0,1\}^{I} are functions I{0,1}I \to \{0,1\}, and the product topology is the topology of pointwise convergence.

Define a sequence (a n)(a_n) in I II^{I} with a n(x)a_n(x) being the nth digit in the binary expansion of xx (we make the convention that binary expansions do not end in sequences of 11s). Let a(a n k)a \coloneqq (a_{n_k}) be a subsequence and define p aIp_a \in I by the binary expansion that has a 00 in the n kn_kth position if kk is even and a 11 if kk is odd (and, for definiteness and to ensure that this fits with our convention, define it to be 00 elsewhere). Then the projection of (a n k)(a_{n_k}) at the p ap_ath coordinate is 1,0,1,0,...1, 0, 1,0,... which is not convergent. Hence (a n k)(a_{n_k}) is not convergent.

(Basically that’s the diagonal trick of Cantor's theorem).

However, as {0,1} I\{0,1\}^I is compact, aa has a convergent subnet. An explicit construction of a convergent subset, given a cluster point aa, is as follows. A function a:I{0,1}a \colon I \to \{0,1\} is a cluster point of (a n)(a_n) if, for any p 1,,p nIp_1, \dots, p_n \in I the set

A(p 1,,p n){k:a k(p i)=a(p i)i} A(p_1,\dots,p_n) \coloneqq \{k \in \mathbb{R} : a_k(p_i) = a(p_i) \forall i\}

is infinite. We index our subnet by the family of finite subsets of II and define the subnet function (I)\mathcal{F}(I) \to \mathbb{N} to be

{p 1,,p n}minA(p 1,,p n) \{p_1,\dots,p_n\} \mapsto \min A(p_1,\dots,p_n)

This is a convergent subnet.

References

This counterexample is based on the one in item 105 of the book

  • Steen, Seebach: Counterexamples in Topology

Revised on July 30, 2010 00:09:16 by Toby Bartels (98.16.141.59)