nLab sequentially compact topological space

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 $\mathbb{N}$, 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.

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.

Proof

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

$$\pi_k: \prod_j X_j \to X_k$$

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

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

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

   Theorem

   Let $\mathcal{T}_1$ and $\mathcal{T}_2$ be two topologies on a set $X$ such that:

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

   then $\mathcal{T}_1 = \mathcal{T}_2$.

   Proof

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

2. The image of a sequentially compact space $X$ under a continuous map $f: X \to Y$ is also sequentially compact. For suppose $y_n$ is a sequence in $f(X)$, say $y_n = f(x_n)$. Then $x_n$ has a convergent subsequence $x_{n_j}$, converging to $x$ say, and by continuity $y_{n_j} = f(x_{n_j})$ converges to $f(x)$.

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:

• Examples of a sequentially compact spaces which are not compact are given in in (Buskes-Rooij 97, example 13.5, Patty 08, chapter 4, example 13, Vermeeren 10, prop. 18). The long line is such an example.

• An example of a compact topological space which is not sequentially compact is given in (Steen-Seebach 70, item 105), recalled at Vermeeren 10, prop. 17. See at compact space – Compact spaces which are not sequentially compact.

Example

(a compact space which is not sequentially compact)

Consider the following uncountable power of the discrete space $2$, namely the product topological space (with its Tychonoff topology)

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

of copies of the discrete space on two elements, indexed by functions $f: \mathbb{N} \to \{0, 1\}$. Since $Disc(\{0,1\})$ is a finite discrete topological space it is clearly compact. Therefore the Tychonoff theorem says that also $X$ is compact.

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

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

i.e., define $x_n$ to have coordinate at $f: \mathbb{N} \to 2$ given by $(x_n)_f = f(n)$. We claim this sequence has no subsequence that converges in $X$, so that $X$ 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 $(x_{(n_k)})_{k \in \mathbb{N}}$ converges to some $x \in X$. Choose any $f: \mathbb{N} \to 2$ that is not eventually constant on the subsequence $(n_k)_{k: \mathbb{N}}$; for example, define $f: \mathbb{N} \to 2$ by $f(n_k) = k\; mod\; 2$, else $f(n) = 0$ if $n$ does not appear in the subsequence. Consider the open set $U_f = \{x_f\} \times \prod_{g: g \neq f} \{0, 1\}$, which is an open neighborhood of $x$. In order to have $x_{n_k} \in U_f$ for all $k \geq k_0$, we’d have to have $f(n_k) = x_f$ for all $k \geq k_0$, in other words $f$ would be eventually constant on the subsequence $n_k$. Contradiction.

Related concepts

• compact topological space

• countably compact topological space

• paracompact topological space

• locally compact topological space

• compactly generated topological space

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

• Wikipedia, Sequentially compact space