nLab sequentially compact topological space




topology (point-set topology, point-free topology)

see also differential topology, algebraic topology, functional analysis and topological homotopy theory


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Universal constructions

Extra stuff, structure, properties


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topological homotopy theory



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 \mathbb{N}, and the one-point compactification {*}\mathbb{N} \cup \{*\} 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.



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


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 j}\{X_j\} be a countable family of sequentially compact spaces. Let

π k: jX jX k\pi_k: \prod_j X_j \to X_k

denote the k thk^{th} projection, and let (x(n))(x(n)) be a sequence in X j\prod X_j. We build a sequence of nested subsequences S j=x j(n)S_j = x_j(n) as follows. For j=1j = 1, choose a subsequence S 1S_1 of the given sequence (x n)(x_n) such that x 1(1)=x(1)x_1(1) = x(1) and such that π 1(S 1)\pi_1(S_1) converges to a point y 1X 1y_1 \in X_1. Then, supposing that S jS_j has been constructed, choose S j+1S_{j+1} to be a subsequence of S jS_j that starts the same way as S jS_j for the first jj steps,

x j+1(1)=x j(1),,x j+1(j)=x j(j),x_{j+1}(1) = x_j(1), \ldots, x_{j+1}(j) = x_j(j),

and such that π j+1(S j+1)\pi_{j+1}(S_{j+1}) converges to y j+1X j+1y_{j+1} \in X_{j+1}. By the axiom of dependent choice, we may suppose these choices have been made for all jj, and then we form the diagonal sequence SS given by x j(j)x_j(j). This is a subsequence of each S jS_j, and so π j(S)\pi_j(S) converges to y jy_j for every jj, hence SS converges to the point (y j) jX j(y_j) \in \prod_j X_j, and we have produced the desired convergent subsequence.

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

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


    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.


    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.

  2. 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:


(a compact space which is not sequentially compact)

Consider the following uncountable power of the discrete space 22, namely the product topological space (with its Tychonoff topology)

Xf:2Disc({0,1}) X \coloneqq \underset{f: \mathbb{N} \to 2}{\prod} Disc(\{0,1\})

of copies of the discrete space on two elements, indexed by functions f:{0,1}f: \mathbb{N} \to \{0, 1\}. 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.

Let (x n) n:(x_n)_{n: \mathbb{N}} be the sequence in XX given by the double-dual embedding

2 2 ,\mathbb{N} \to 2^{2^\mathbb{N}},

i.e., define x nx_n to have coordinate at f:2f: \mathbb{N} \to 2 given by (x n) f=f(n)(x_n)_f = f(n). We claim this sequence has no subsequence that converges in XX, so that XX is not sequentially compact. We will argue by contradiction. (This refutation by contradiction, refuting sequential compactness, is constructively valid. The argument above that affirms compactness, via the Tychonoff theorem, is not constructive.)

Suppose instead some subsequence (x (n k)) k(x_{(n_k)})_{k \in \mathbb{N}} converges to some xXx \in X. Choose any f:2f: \mathbb{N} \to 2 that is not eventually constant on the subsequence (n k) k:(n_k)_{k: \mathbb{N}}; for example, define f:2f: \mathbb{N} \to 2 by f(n k)=kmod2f(n_k) = k\; mod\; 2, else f(n)=0f(n) = 0 if nn does not appear in the subsequence. Consider the open set U f={x f}× g:gf{0,1}U_f = \{x_f\} \times \prod_{g: g \neq f} \{0, 1\}, which is an open neighborhood of xx. In order to have x n kU fx_{n_k} \in U_f for all kk 0k \geq k_0, we’d have to have f(n k)=x ff(n_k) = x_f for all kk 0k \geq k_0, in other words ff would be eventually constant on the subsequence n kn_k. Contradiction.


  • 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

Last revised on December 21, 2020 at 01:04:19. See the history of this page for a list of all contributions to it.