analysis (differential/integral calculus, functional analysis, topology)
metric space, normed vector space
open ball, open subset, neighbourhood
convergence, limit of a sequence
compactness, sequential compactness
continuous metric space valued function on compact metric space is uniformly continuous
…
…
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
The classical Heine–Borel theorem identifies those topological subspaces of Cartesian spaces ($\mathbb{R}^n$s) that are compact in terms of simpler properties. A generalisation applies to all metric spaces and even to uniform spaces.
This is the classical theorem:
Let $S$ be a topological subspace of a Cartesian space ($S \subset \mathbb{R}^n$). Then $S$ is a compact topological space (with the induced topology) precisely if it is closed and bounded in $\mathbb{R}^n$.
It's easy to prove that $S$ is closed precisely if it is a complete metric space as with the induced metric, and similarly $S$ is bounded precisely if it is totally bounded. This gives the next version:
Let $S$ be a topological subspace of a Cartesian space ($S \subset \mathbb{R}^n$). Then $S$ is a compact topological space (with the induced topology) precisely if it is complete and totally bounded (with the induced metric).
This refers entirely to $S$ as a metric space in its own right. In fact it holds much more generally than for subspaces of a cartesian space:
Let $S$ be a metric space. Then $S$ is compact precisely if it is complete and totally bounded.
This theorem refers only to uniform properties of $S$, and in fact a further generalistion is true:
Let $S$ be a uniform space. Then $S$ is compact precisely if it is complete and totally bounded.
The hard part is proving that a complete totally bounded space is compact; the converse is easy.
We could also try to generalise Theorem to subspaces of other metric spaces, but this fails: every compact subspace of a metric space is closed and bounded (which is the easy direction), but not conversely.
Another variant of these statements is:
In the old days, one called a closed and bounded interval in the real line ‘compact’; once closedness and boundedness were generalised from intervals to arbitrary subsets (of the real line), the definition of ‘compact’ also generalised. The content of Theorem , then, is that this condition is equivalent to the modern definition of ‘compact’ using open covers, and indeed the modern definition was only derived afterwards as a name for the conclusion of the theorem.
In constructive mathematics, one sees several definitions of ‘compact’, which may make the theorem provable, refutable, or undecidable in various constructive systems. Using the open-cover definition of ‘compact’, Theorems and are equivalent to the fan theorem (and so hold in intuitionism), but Theorems and are stronger and Brouwer could not prove them, leading him to define ‘compact’ (for a metric space) to mean complete and totally bounded. In other literature, one sometimes sees the abbreviation ‘CTB’ used instead. In Russian constructivism, already Theorems and can be refuted using the open-cover definition, but CTB spaces are still important.
In locale theory and other approaches to pointless topology, the open-cover definition of ‘compact’ is clearly correct, and the failure of CTB spaces to be compact (constructively) may be seen as a consequence of working with points. Already in Bishop's weak system of constructivism, every CTB metric space $X$ gives rise to a compact locale, which classically (assuming excluded middle and dependent choice) is the locale of open subsets of $X$ but constructively requires a more nuanced construction; see Vickers.
In dependent type theory, the open-cover definition of ‘compact’ holds as well, but one has to use the inductive covers higher inductive type instead of the usual pointwise definitions of covers (UFP13, §11.5).
For definiteness, we restate the versions which we prove:
(closed interval is compact)
For any $a \lt b \in \mathbb{R}$ the closed interval
regarded with its subspace topology is a compact topological space.
Since all the closed intervals are homeomorphic it is sufficient to show the statement for $[0,1]$. Hence let $\{U_i \subset [0,1]\}_{i \in I}$ be an open cover. We need to show that it has an open subcover.
Say that an element $x \in [0,1]$ is admissible if the closed sub-interval $[0,x]$ is covered by finitely many of the $U_i$. In this terminlogy, what we need to show is that $1$ is admissible.
Observe from the definition that
0 is admissible,
if $y \lt x \in [0,1]$ and $x$ is admissible, then also $y$ is admissible.
This means that the set of admissible $x$ forms either an open interval $[0,g)$ or a closed interval $[0,g]$, for some $g \in [0,1]$. We need to show that the latter is true, and for $g = 1$. We do so by observing that the alternatives lead to contradictions:
Assume that the set of admissible values were an open interval $[0,g)$. By assumption there would be a finite subset $J \subset I$ such that $\{U_i \subset [0,1] \}_{i \in J \subset I}$ were a finite open cover of $[0,g)$. Accordingly, since there is some $i_g \in I$ such that $g \in U_{i_g}$, the union $\{U_i\}_{i \in J } \sqcup \{U_{i_g}\}$ were a finite cover of the closed interval $[0,g]$, contradicting the assumption that $g$ itself is not admissible (since it is not contained in $[0,g)$).
Assume that the set of admissible values were a closed interval $[0,g]$ for $g \lt 1$. By assumption there would then be a finite set $J \subset I$ such that $\{U_i \subset [0,1]\}_{i \in J \subset I}$ were a finite cover of $[0,g]$. Hence there would be an index $i_g \in J$ such that $g \in U_{i_g}$. But then by the nature of open subsets in the Euclidean space $\mathbb{R}$, this $U_{i_g}$ would also contain an open ball $B^\circ_g(\epsilon) = (g-\epsilon, g + \epsilon)$. This would mean that the set of admissible values includes the open interval $[0,g+ \epsilon)$, contradicting the assumption.
This gives a proof by contradiction.
(Heine-Borel theorem classically)
For $n \in \mathbb{N}$, regard $\mathbb{R}^n$ as the $n$-dimensional Euclidean space, regarded as a topological space via its metric topology.
Then in classical mathematics for a topological subspace $S \subset \mathbb{R}^n$ the following are equivalent:
First consider a subset $S \subset \mathbb{R}^n$ which is closed and bounded. We need to show that regarded as a topological subspace it is compact.
The assumption that $S$ is bounded by (hence contained in) some open ball $B^\circ_x(\epsilon)$ in $\mathbb{R}^n$ implies that it is contained in $\{ (x_i)_{i = 1}^n \in \mathbb{R}^n \,\vert\, -\epsilon \leq x_i \leq \epsilon \}$. This topological subspace is homeomorphic to the $n$-cube $[-\epsilon, \epsilon]^n$. Since the closed interval $[-\epsilon, \epsilon]$ is compact by lemma , the Tychonoff theorem implies that this $n$-cube is compact.
Since subsets are closed in a closed subspace precisely if they are closed in the ambient space the closed subset $S \subset \mathbb{R}^n$ is still closed as a subset $S \subset [-\epsilon, \epsilon]^n$. Since closed subspaces of compact spaces are compact this implies that $S$ is compact.
Conversely, assume that $S \subset \mathbb{R}^n$ is a compact subspace. We need to show that it is closed and bounded.
The first statement follows since the Euclidean space $\mathbb{R}^n$ is Hausdorff and since compact subspaces of Hausdorff spaces are closed.
Hence what remains is to show that $S$ is bounded.
To that end, choose any positive real number $\epsilon \in \mathbb{R}_{\gt 0}$ and consider the open cover of all of $\mathbb{R}^n$ by the open n-cubes
for n-tuples of integers $(k_1, k_2 , \cdots, k_n ) \in \mathbb{Z}^n$. The restrictions of these to $S$ hence form an open cover of the subspace $S$. By the assumption that $S$ is compact, there is then a finite subset of $n$-tuples of integers such that the corresponding $n$-cubes still cover $S$. But the union of any finite number of bounded closed $n$-cubes in $\mathbb{R}^n$ is clearly a bounded subset, and hence so is $S$.
According to Wikipedia, the theorem was first proved by Pierre Cousin in 1895. It is named after Eduard Heine (who used it but did not prove it) and Émile Borel (who proved a limited version of it), an instance of Baez's law.
A proof is spelled out for instance at
On constructing a compact locale from a CTB metric space:
On the Heine-Borel theorem in dependent type theory using inductive covers, see section 11.5 of:
Last revised on November 23, 2022 at 18:27:33. See the history of this page for a list of all contributions to it.