I am not asking for properties equivalent to compactness, but for those that better capture the motivation for compactness, i.e. that explain why compactness is talked about so much.

The way I see it compactness captures (in the form of sequential compactness) a property of closed and bounded subsets in an euclidean space: every sequence has a converging subsequence. Addressing this property first came Bolzano's theorem (1817) and then results on spaces of functions like the Arzelà–Ascoli theorem. Simultaneously there was the study of the continuum and proof that a continuous function defined on a closed and bounded interval is uniformly continuous (1870), which led to the modern definition of compactness (in terms of open covers), which is closely related to sequential compactness but suitable for general topology.

My question would be the following:

If you had to write a short list of properties (say no more than five) that you'd consider the most important consequences of compactness (in any context, from general topology to a more specific setting), a sort of guideline of what compactness means, which ones would you choose?

  • $\begingroup$ Compactness is the key to existence results (e.g. in PDE). $\endgroup$
    – Thomas
    Jan 2, 2018 at 13:17
  • 2
    $\begingroup$ a compact set behaves in some ways as a finite set, this is the reason of it name: compact. It more important property becomes from it definition, that is, that for any open cover exists a finite open cover. $\endgroup$
    – Masacroso
    Jan 2, 2018 at 13:20
  • 1
    $\begingroup$ In my opinion, compactness is the weapon to tackle infinity with only the knowledge of the finite case. Think of finite intersection property. $\endgroup$ Jan 2, 2018 at 13:40
  • 1
    $\begingroup$ Maybe you take a look at the following question: math.stackexchange.com/questions/371928/… $\endgroup$ Jan 2, 2018 at 14:36

4 Answers 4


The first elementary but crucial property, I would think of when talking about consequences of compactness, would be the following:

Proposition $1$. Let $X$ be a compact topological set and let $(K_i)_{i\in I}$ be a family of closed sets of $X$, then: $$\bigcap_{i\in I}K_i\neq\varnothing\iff\left(\forall J\subset I,\#J<+\infty\Rightarrow\bigcap_{j\in J}K_j\neq\varnothing\right).$$

Proof. The direct implication is obvious.

For the reverse one let us proceed by the contrapositive and assume that one has: $$\bigcap_{i\in I}K_i=\varnothing.$$ Then, taking the complementary, one gets: $$\bigcup_{i\in I}X\setminus K_i=X.$$ Since $X$ is compact, there exists $J\subset I$ finite such that: $$\bigcup_{j\in J}X\setminus K_j=X.$$ Whence the result taking again the complementary. $\Box$

Remark 1. This proposition is frequently applied when $(K_i)_{i\in I}$ is a decreasing sequence of nonempty compact sets of a topological compact set, not necessarily compact itself.

Here is an explicit example where proposition $1$ is used. It is also a good example of what one can expect from compactness assumptions.

Theorem $1$. Let $E$ be a real vector space endowed with a dot product, let $K$ be a nonempty compact convex set of $E$ and let $G$ be a compact subgroup of $\textrm{GL}(E)$. Assume that $G$ stabilises $K$, then there exists $x\in K$ such that for all $g\in G$, $g(x)=x$.

The key lemma of theorem $1$ is the following:

Lemma $1$. Let $T$ be a continuous linear map of $E$ and assume that $T(K)\subset K$, then there exists $x\in K$ such that $T(x)=x$.

Proof. Let $x_0\in K$ and let define the following sequence: $$x_{n+1}=\frac{1}{n+1}\sum_{k=0}^nT^k(x_0)\in K,$$ using compactness of $K$, one can assume that $(x_n)_n$ converges toward $x \in K$. To conclude, notice that: $$\|T(x_n)-x_n\|=\frac{\|T^{n+1}(x_0)-x_0\|}{n+1}\leqslant\frac{\textrm{diam}(K)}{n+1}.$$ Therefore, taking $n\to+\infty$ shows that $x$ is a fixed point of $T$. Whence the result. $\Box$

Then, using proposition $1$ applied to: $$K_u:=\{x\in K\textrm{ s.t. }u(x)=x\},$$ one sees that theorem $1$ is true if and only if lemma $1$ holds for a finite number of continuous linear maps. The rough sketch of a proof of theorem $1$ does not capture why the compactness of $G$ is important, but it is.

This result is called Kakutani's fixed point and have deep applications. Here is a modest sample:

Proposition $2.$ Let $G$ be a compact subgroup of $\textrm{GL}_n(\mathbb{R})$, then $G$ is conjugated to a subgroup of $O(n)$.

Remark $2$. The proof uses the fact that the convex hull of a compact set is again compact, which follows from Carathéodory's theorem.

Perhaps, more substantial is the following:

Theorem $2$. Let $G$ be a compact topological group, then there exists a unique Borel probability measure $\mu$ on $G$ such that for all $g\in G$ and all Borel measurable subset $A$ of $G$, one has: $$\mu(gA)=\mu(A).$$ In particular, for all measurable map $f\colon G\rightarrow\mathbb{R}$ and all $g\in G$, one has: $$\int_Gf(gx)\,\mathrm{d}\mu(x)=\int_Gf(x)\,\mathrm{d}\mu(x).$$

The existence of such measures is of importance in Riemannian geometry.

Proposition $3.$ Let $G$ be a compact Lie group, then $G$ can be endowed with a bi-invariant Riemannian metric i.e. such that left and right translations are isometries.

Proof. Let $\langle\cdot,\cdot\rangle$ be a dot product on $\mathfrak{g}$, let $\mu$ be the measure of theorem $2$ and let define: $$\forall x,y\in\mathfrak{g},\langle x,y\rangle_G:=\int_{g\in G}\langle\textrm{Ad}(g)x,\textrm{Ad}(g)y\rangle\,\mathrm{d}\mu(g),$$ then $\langle\cdot,\cdot\rangle_G$ is a dot product on $\mathfrak{g}$ which is $\textrm{Ad}\colon G\rightarrow\textrm{GL}(\mathfrak{g})$ invariant and $h\in\Gamma(TG\otimes TG)$ defined by: $$h_g:={L_g}^*\langle\cdot,\cdot\rangle_G$$ is a bi-invariant metric on $G$. Whence the result. $\Box$

Remark 3. In fact, the set of bi-invariant metric on $G$ is in bijective correspondence with the set of $\textrm{Ad}$-invariant dot product on $\mathfrak{g}$.

From proposition $3$, it is straightforward to deduce the following:

Theorem $3$. The exponential map of a compact Lie group is surjective.

Proof. Let $G$ be a compact Lie group, by proposition $3$, let $h$ be a bi-invariant metric on $G$ and let $\nabla$ be the associated Levi-Civita connection, then using the Koszul's formula, one has: $$\forall X,Y\in{}^G\Gamma(TG),\nabla_XY=\frac{1}{2}[X,Y].$$ From there, it is easy to see that the geodesics starting from the identity element of $G$ are the one parameter subgroups of $G$. Therefore, the Lie theoretical exponential map of $G$ coincides with the Riemannian exponential of $(G,h)$. Whence the result from Hopf-Rinow theorem. $\Box$

Remark 4. In the proof, I used that an $\textrm{Ad}$-invariant bilinear form $B$ on $\mathfrak{g}$ is $\textrm{ad}$-alternate, namely: $$B([X,Y],Z)=B(X,[Y,Z]).$$

Example $1$. The matrix exponential map from $\textrm{SO}(n)$ to the set of skew-symmetric matrices is surjective, since for matrix Lie groups, the exponential map is the usual one.


Let $f$ be lower semicontinuous and coercive, i.e., $f(x)\to +\infty$ whenever $\|x\|\to+\infty$. Then $f$ has a minimizer.

This beautiful result, in Euclidean spaces, generalizes the Weierstrass theorem and lies at the heart of optimization: existence of minimizers! At its core is Bolzano-Weierstrass and thus compactness.

In infinite-dimensional Hilbert space, we require that $f$ be weakly lower semicontinuous, so weak sequential compactness (Eberlein-Smulyan!) is the key.


Continuous functions have maximum and minimum. Bounded sequences have converging subsequences.

These are the main motivations to consider compactness in analysis.


My short list has one item: the finite intersection property.

This property can be generalized to partially ordered sets, for example $\langle P, \leq, \bot \rangle$ with least element $\bot$. The way to do this is to define a partial function $\wedge$ that, maps a nonempty subset of $P$, say $A$, to the greatest lower bound of $A$, provided a greatest lower bound of $A$ exists. We can say that $A$ satisfies the finite meet property if, for all $B \subseteq A$ with $\wedge B = \bot$ there exists a finite $F \subseteq B$ with $\wedge F = \bot$. If $A$ satisfies the finite meet property then, for all $A \in A$ there exists an $m \in A$ satisfying both $\bot < m \leq a$ and, for all $x \in A$, if $x \leq m$ then $x = \bot$ or $x = m$.

The the finite meet property can be generalized. The nonempty set $A \subseteq P$ satisfies the eventual finite meet property if, for all $a_{1} \in A$ with $\bot < a_{1}$, there exists an $a_{0} \in A$ satisfying both $\bot < a_{0} \leq a_{1}$ and $\{ a \in A \colon a \leq a_{0} \}$ satisfies the finite meet property. The usefulness of this property is that $A$ satisfies the eventual finite meet property iff, for all $a \in A$ with $a > \bot$, there exists a minimal $m \in A$ with $m \leq a$.

Example 1. Tet $\mathcal{P}$ be the collection of the following sets:

  • For $m \in \mathbb{N}$ the set $\{ n \in \mathbb{N} \colon n \geq m \} $.
  • If $p$ is prime the set $\{ p \} $.

This collection of sets satisfies the eventual finite meet property but not the finite meet property.

Example 2. The Krein-Milman theorem says a compact convex set in a locally convex Hausdorff topological vector space is the closed convex hull of its extreme points. Let $B$ be the closed unit ball in an infinite dimensional Hilbert space. The extreme points of $B$ is the set of unit vectors. The set $B$ is the convex hull of its extreme points. Thus $B$ satisfies the conclusion of the Krein-Milman theorem but not the hypothesis. This suggests the hypothesis of compactness can be weakened. Check that the collection of extreme subsets of $B$ satisfies the eventual extreme subset property.

  • $\begingroup$ In example 2, do you mean compact and convex, rather than “closed compact”? The space is Hausdorff. $\endgroup$ Jan 3, 2018 at 0:41
  • $\begingroup$ @TheoreticalEconomist Thanks, for the correction. $\endgroup$
    – Jay
    Jan 4, 2018 at 13:04

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