topology (point-set topology, point-free topology)
see also differential topology, algebraic topology, functional analysis and topological homotopy theory
Basic concepts
fiber space, space attachment
Extra stuff, structure, properties
Kolmogorov space, Hausdorff space, regular space, normal space
sequentially compact, countably compact, locally compact, sigma-compact, paracompact, countably paracompact, strongly compact
Examples
Basic statements
closed subspaces of compact Hausdorff spaces are equivalently compact subspaces
open subspaces of compact Hausdorff spaces are locally compact
compact spaces equivalently have converging subnet of every net
continuous metric space valued function on compact metric space is uniformly continuous
paracompact Hausdorff spaces equivalently admit subordinate partitions of unity
injective proper maps to locally compact spaces are equivalently the closed embeddings
locally compact and second-countable spaces are sigma-compact
Theorems
Analysis Theorems
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.
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.
For a metric space, the notions of sequential compactness and compactness coincide. See at sequentially compact metric spaces are equivalently compact metric spaces.
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.
A countable product of sequentially compact spaces is again sequentially compact.
Let $\{X_k\}$ be a countable family of sequentially compact spaces. Let $(a_l)$ be a sequence in $\prod X_k$. For each $m$ we recursively define an infinite subset $A_m \subseteq A_{m-1} \subseteq \mathbb{N}$ with the property that the sequence $(a_l)_{l \in A_m}$ converges when projected down to $\prod_{k=1}^m X_k$. Let $l_m = \min\{A_l\}$. Consider the sequence $(a_{l_m})$. For each $k$, we choose a limit $x_k$ of the projection of $(a_l)_{l \in A_k}$ to $X_k$. Let $x = (x_k) \in \prod X_k$. Let $U$ be a neighbourhood of $x$. Then there is some $n \in \mathbb{N}$ and neighbourhood $U_n \subseteq \prod_{k=1}^n X_k$ of $(x_k)_{k=1}^n$ such that $U$ contains the preimage of $U_n$. For $m \ge n$, the sequence $(l_m)$ is contained in $A_n$ and so the image of $(a_{l_m})$ converges to $(x_k)_{k=1}^n$. Hence there is some $r$ such that for $m \ge r$, the projection of $a_{l_m}$ lies in $U_n$. Hence for $m \ge r$, $a_{l_m} \in U$. Thus $(a_{l_m})$ converges to $(x_k)$ and so $\prod X_k$ is sequentially compact.
This shows that the example of a compact space that is not sequentially compact is about as simple as can be.
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 $\mathcal{T}_1$ and $\mathcal{T}_2$ be two topologies on a set $X$ such that:
then $\mathcal{T}_1 = \mathcal{T}_2$.
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.
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 $f(x)$.
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).
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).
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.
(a compact space which is not sequentially compact)
Consider the product topological space (with its Tychonoff topology)
of copies of the discrete space on two elements, indexed by the points in the half-open interval. Since $Disc(\{0,1\})$ is a finite discrete topological space it is clearly compact. Therefore the Tychonoff theorem says that also $X$ is compact.
But here is an instance of a sequence $x_{(-)}$ in $X$ which does not have a convergent sub-sequence:
By the nature of the product, an element $x \in X$ is a tuple of elements $\pi_r(x) \in \{0,1\}$ for $r \in [0,1)$. Now for $n \in \mathbb{N}$ define $x_n$ by
Suppose this sequence $(x_n)_{n \in \mathbb{N}}$ had a subsequence $(x_{(n_k)})_{k \in \mathbb{N}}$, converging to some $x_\infty \in X$, hence that for every open neighbourhood $\{x_\infty\} \subset U \subset X$ there were a $k_0 \in \mathbb{N}$ such that $x_{n_{k \geq k_0}} \in U$.
Considering then for each $r \in [0,1)$ the Tychonoff-open subset
this would imply that there were $n_0$ such that
hence that the binary expansion of $r$ had the same digit in position $k_n$ for all $n \geq n_0$. This is clearly not the case for all $r$ (if it is true for some $r$, just change one of these digits to obtain an $r'$ for which it is not), and hence we have a proof by 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 May 15, 2017 at 09:09:43. See the history of this page for a list of all contributions to it.