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For a strictly distance-decreasing map of a finite-dimensional Euclidean space, one bounded forward orbit suffices for existence of a fixed point. Its closure is compact by the Heine-Borel theorem and invariant by continuity, so strict distance decrease on a compact metric space applies. Strict distance decrease gives uniqueness throughout the whole space, not merely the orbit closure.

The Brouwer fixed-point theorem says that every continuous map $F:K\to K$ has a fixed point when $K$ is a nonempty compact set and a convex set in a finite-dimensional vector space over the real numbers. Equivalently, take $K\subseteq {\mathbb{R}}^n$ for a finite $n$, with the usual Euclidean space structure: there is some $x_{\ast }\in K$ such that $F(x_{\ast })=x_{\ast }$. The conclusion guarantees at least one fixed point, without asserting uniqueness or telling us how quickly to find one. ^[ref]

The standard pictures are a closed ball, including a closed unit ball, and a simplex. A filled disk qualifies; its bounding circle alone does not. A probability simplex also qualifies even though its dimension is smaller than that of its ambient Euclidean space. The hypotheses have concrete meanings: being a convex set keeps every line segment between allowed points inside $K$, and compactness prevents escape or loss of a limiting point. In Euclidean space, the Heine-Borel theorem identifies compact sets with closed sets that are bounded sets. Crucially, every output must belong to the same $K$ as the input.

For a fully computable example, let $D=\{(u,v):u^2+v^2\le 1\}$ be the filled unit disk. Apply a counterclockwise quarter-turn rotation, shrink by $0.4$, and add a translation:This affine map is a continuous map. Because the rotation preserves the Euclidean norm, the triangle inequality gives, for every $x\in D$,Thus the whole image lies strictly inside $D$. More precisely, it is a disk of radius $0.4$ centred at $(0.3,0.1)$. This containment check is what makes the Brouwer fixed-point theorem applicable; a formula being a continuous map by itself would not suffice.

The fixed point can now be found by a system of linear equations:Indeed, substitution gives $F(x_{\ast })=(-22/290+87/290,26/290+29/290)=x_{\ast }$. Its Euclidean norm is $\sqrt{290}/58\approx 0.293610$, safely inside $D$. The figure shows arrows from sample inputs to their outputs; an arrow disappears at the fixed point because the displacement $F(x)-x$ vanishes there.

Repeated iterations of a map happen to succeed for this example. Starting at $x_0=(1,0)$ and setting $x_{k+1}=F(x_k)$ gives $x_1=(0.3,0.5)$, $x_2=(0.1,0.22)$, and $x_3=(0.212,0.14)$. Subtracting the fixed point equation yields $x_{k+1}-x_{\ast }=0.4R_{90}(x_k-x_{\ast })$, so the Euclidean distance to $x_{\ast }$ shrinks by exactly $0.4$ each step. This extra fact follows from the contraction mapping property and the Banach fixed-point theorem. It is additional information beyond the Brouwer fixed-point theorem.

The distinction becomes visible with two simpler continuous maps of the same disk. The pure rotation $G(u,v)=(-v,u)$ has exactly one fixed point, $(0,0)$, since $u=-v$ and $v=u$ force both to vanish. Yet iterations of a map starting at $(1,0)$ here cycle through $(0,1)$, $(-1,0)$, and $(0,-1)$ forever. Their Euclidean distance from $(0,0)$ stays one, so there is no convergence in a metric space. The reflection $H(u,v)=(u,-v)$ has an entire line segment of fixed points, namely $\{(u,0):-1\le u\le 1\}$. Starting off that line segment produces an alternating pair. Existence, uniqueness, and convergence of iterations of a map are separate questions.

Removing hypotheses changes the answer. Restrict the same quarter-turn rotation to the circle $S^1=\{(u,v):u^2+v^2=1\}$. It remains a continuous map into that compact set, but its only possible fixed point, $(0,0)$, has been removed. The circle is not a convex set. Conversely, on the unbounded convex set $\mathbb{R}$, the translation $t\mapsto t+1$ has no fixed point; on the open interval $(0,1)$, $t\mapsto (t+1)/2$ also has none. Neither domain is a compact set. The finite-dimensional closed ball statement should likewise not be extended to arbitrary infinite-dimensional vector spaces, where a closed ball need not be a compact set.

The one-dimensional version explains the connection to the intermediate value theorem. For a continuous map $f:[a,b]\to [a,b]$, put $h(t)=f(t)-t$. Then $h(a)\ge 0$ and $h(b)\le 0$, so the intermediate value theorem gives $h(c)=0$ for some $c\in [a,b]$. In higher dimensions, the Brouwer fixed-point theorem supplies a corresponding existence principle when this scalar sign argument is unavailable. Its practical value is ensuring that a proposed equation $x=F(x)$ has a solution once a suitable domain is found; it supplies no general promise of a fast algorithm.

A useful application is positive matrix eigenvector from simplex normalization: rescale the output of a strictly positive matrix to make its entries sum to one. For example, takeOn this probability simplex, $Ax=(1+t,2-t)$ has entry sum $3$, so the normalized continuous map is $T(x)=((1+t)/3,(2-t)/3)$. Both entries are positive and sum to one, keeping the image inside the probability simplex. Its fixed point equation is $t=(1+t)/3$, hence $t=1/2$. Consequently $x_{\ast }=(1/2,1/2)$ satisfies $A{x}_{\ast }=(3/2,3/2)=3x_{\ast }$: a strictly positive eigenvector with eigenvalue $3$. In larger dimensions, the Brouwer fixed-point theorem guarantees the corresponding positive eigenvector without requiring us to solve a system of linear equations in advance. The fixed point expresses preservation of relative weights, rather than preservation of their unnormalized size.

Proofs can be approached through the no-retraction theorem or Sperner's lemma. In the latter approach, increasingly fine triangulations locate approximate fixed points; compactness supplies a convergent subsequence, and continuous maps turn its limit into an exact fixed point. ^[ref]

A primary historical reference is L. E. J. Brouwer's 1911 paper Über Abbildung von Mannigfaltigkeiten, whose digitized original is hosted by Zenodo under DOI 10.1007/BF01456812. This gives a direct route to the early literature behind the Brouwer fixed-point theorem. ^[ref]

A bounded closed interval includes both of its endpoints:If $a=b$, it consists of one point. Its length is $b-a$. In contrast, the open interval $(a,b)$ excludes both endpoints; half-open real intervals include just one.

Including the endpoints has a useful consequence for limits: a convergent sequence whose terms lie in $[a,b]$ has its limit of a sequence in $[a,b]$ too. Finite endpoints also make this a bounded set, so it is a compact set by the Heine-Borel theorem. These two properties support both the nested interval theorem and the extreme value theorem.

A compact group is a topological group whose underlying topological space is a compact Hausdorff space. The circle group is a compact group: the unit circle is a closed bounded subset of the complex plane, so the Heine-Borel theorem gives compactness. The additive group $(\mathbb{R},+)$ is a locally compact group and a Hausdorff space, but is not a compact group: the open cover $\{(-n,n):n\ge 1\}$ has no finite subcover. A finite group with the discrete topology is another compact group; compactness does not require a smooth manifold structure.

The great advantage of a compact group is that one can average over every element without choosing a preferred origin. Precisely, every locally compact group that is a Hausdorff space admits a nonzero left Haar measure $\mu$, a regular Borel measure finite on compact sets, unique up to a positive scalar. Left invariance means $\mu (gA)=\mu (A)$ for every Borel set $A$ and every $g\in G$. There is also a right Haar measure; in general left and right Haar measures can differ. On a compact group, $0<\mu (G)<\infty$, and normalizing to $\mu (G)=1$ gives a unique regular Borel probability measure invariant under both left and right translation. See Tornier, Haar Measures, Theorem 2.2 for existence and uniqueness, and Ram's notes on left and right Haar measure for the distinction.

For a finite group $G$, normalized Haar measure assigns each element mass $1/|G|$, so integration is simply $|G{|}^{-1}{\sum }_{g\in G}f(g)$. For the circle group, the corresponding integral is uniform angle averaging, computed below. On $(\mathbb{R},+)$, Lebesgue measure is a Haar measure, but its total mass is infinite: no positive scalar multiple makes it a probability measure. Thus the existence of Haar measure does not make every topological group a compact group.

Historically, Alfréd Haar introduced the general invariant-measure construction in Der Massbegriff in der Theorie der kontinuierlichen Gruppen, Annals of Mathematics 34 (1933), pp. 147-169. This bibliographic identification is also recorded in André Weil's archived seminar notes, Mesure de Haar, reference Haa33. The conceptual step is to let group symmetry itself determine the weights used in integration, rather than relying on coordinates in an ambient Euclidean space.

A compact interval in the real numbers is either the empty set or a closed bounded interval $[a,b]$ with $a\le b$. Its compactness follows from the Heine-Borel theorem.

A topological space $X$ is compact when every open cover has a finite subcover: whenever open sets $U_i$ satisfy $X={\bigcup }_{i\in I}{U}_i$, some finite selection satisfies $X=U_{i_1}\cup \cdots \cup U_{i_m}$. A subset $K$ of an ambient topological space is compact when its subspace topology is compact; equivalently, every open cover of $K$ by ambient open sets has a finite subcover. The word every matters: finding one finite subcover of one open cover is not enough to establish compactness. ^[ref]

The closed interval $[0,1]$ is compact by the Heine-Borel theorem. For a concrete open cover, take the three open intervals $(-0.1,0.45)$, $(0.35,0.8)$, and $(0.7,1.1)$. Their union is $(-0.1,1.1)$, so they cover every point of $[0,1]$, including both endpoints. The overlaps have lengths $0.1$ and $0.1$, leaving no gaps. Each member is necessary for this particular open cover: removing them respectively leaves $0$, $0.5$, or $1$ uncovered.

Above, hollow markers denote the excluded endpoints of the covering open intervals, while the solid endpoints belong to the covered closed interval. Below, each displayed open interval is one member of an infinite open cover of $(0,1)$; finitely many rows always leave points near zero uncovered. The rows are vertically separated for readability and all refer to the same real line.

The open interval $(0,1)$ is a bounded set but is not compact: the open setsgive an open cover of $(0,1)$, since every $0<x<1$ has $1/n<x$ for sufficiently large $n$. These are also open sets for the subspace topology on $(0,1)$. Any nonempty finite selection has a largest index $N$, so its union is just $(1/N,1)$ and misses the point $1/(N+1)$, which still belongs to $(0,1)$. Thus no finite subcover exists, however many members we select. The same family is not an open cover of $[0,1]$, since it misses both endpoints.

In a metric space, compactness is equivalent to sequential compactness: every sequence in the subset has a convergent subsequence whose limit of a sequence belongs to that subset. This equivalence does not extend to all topological spaces. For example, $x_n=1/n$ for $n\ge 2$ has limit of a sequence zero in $[0,1]$. The same sequence lies in $(0,1)$, but every subsequence still has limit of a sequence zero on the real line, so no subsequence converges to a point of $(0,1)$. Merely approaching a boundary is harmless when the required point remains available. ^[ref]

Both panels show the sequence $x_n=1/n$ for $n\ge 2$ approaching zero. In the closed interval, zero is an allowed limit of a sequence; in the open interval, the hollow marker identifies the missing point. This single sequence disproves sequential compactness of $(0,1)$; compactness of $[0,1]$ still requires control of every open cover or every sequence, supplied by the Heine-Borel theorem.

An infinite set with the discrete topology similarly fails compactness, using the open cover by all its singletons. Every finite set is compact in any topology: select one covering open set for each point. The empty set is compact because the empty selection is a finite subcover. This open cover definition does not require the topological space to be Hausdorff; in algebraic geometry it is often called quasi-compactness.

The beauty of compactness is that information valid around each point can often be controlled using finitely many regions. A finite subcover lets us combine finitely many local bounds, or choose the smallest of finitely many positive tolerances, to obtain control across the whole compact space. This is why compactness repeatedly turns approximation into existence and why it is so useful in optimization through the extreme value theorem.

Every closed set in a compact space is compact, and a continuous image of a compact space is compact. The latter proof pulls an open cover back by inverse images, selects a finite subcover, and maps forward. This makes compactness useful for proving existence: the extreme value theorem guarantees that a continuous function from a nonempty compact space to the real line attains both a minimum and a maximum. For example, $q(x)=x(1-x)=1/4-(x-1/2{)}^2$ on $[0,1]$ has maximum $1/4$ at $1/2$ and minimum zero at its endpoints. On $(0,1)$ its infimum is still zero but is never attained. ^[ref]

The closed unit ball of a Hilbert space is compact in norm exactly when the space is finite-dimensional. Finite-dimensional compactness follows from the Heine-Borel theorem. In infinite dimension, an orthonormal sequence has pairwise distance $\sqrt{2}$ and no convergent subsequence. Restricting a compact operator that equals a nonzero multiple of the identity on a subspace therefore forces that subspace to be finite-dimensional.

Let $K=\overline{O(x)}$. It is nonempty, closed and bounded in finite-dimensional Euclidean space, hence compact by the Heine-Borel theorem. The strict distance inequality again makes $f$ Lipschitz continuous. If $y\in K$, choose a sequence $y_j\in O(x)$ converging to $y$. Then $f{y}_j\in O(x)$, and continuity gives $f{y}_j\to fy$, so $fy\in K$. This proves $f(K)\subseteq K$.

The restriction to $K$ satisfies strict distance decrease on a compact metric space, so part (b) produces a fixed point there. If a second fixed point existed anywhere in ${\mathbb{R}}^n$, the same strict distance inequality applied to the two points would contradict equality of their distance. Therefore the fixed point is unique in all of ${\mathbb{R}}^n$. This is the bounded-orbit fixed point for a strictly distance-decreasing Euclidean map.

A point $x$ is critical for a smooth map $f:X\to Y$ when $D{f}_x:T_{x}X\to T_{f(x)}Y$ is not surjective. A critical value is the image of a critical point; a regular value has no critical point in its fiber, including the case of an empty fiber.

To prove the regular level set theorem, take $x_0\in f^{-1}(y)$ and write $n=\dim X$, $m=\dim Y$. In local coordinates, surjectivity supplies an invertible $m\times m$ derivative minor; relabel source coordinates so it uses the first $m$ columns. The maphas an invertible derivative. The inverse function theorem makes it a local diffeomorphism. In these coordinates the fiber is obtained by fixing its first $m$ coordinates, so it is locally an open subset of ${\mathbb{R}}^{n-m}$. These smooth charts prove that a nonempty regular fiber is a smooth embedded manifold of dimension $n-m$, with tangent space $\ker D{f}_x$.

For the special orthogonal group, use $F:M_n(\mathbb{R})\to {\mathrm{Sym}}_n(\mathbb{R})$ defined by $F(A)=A{A}^T$. Its derivative is $D{F}_A(H)=H{A}^T+A{H}^T$. At an orthogonal $A$, any symmetric matrix $B$ is obtained by taking $H=BA/2$, so $I$ is a regular value. Therefore $O(n)=F^{-1}(I)$ has dimension $n^2-n(n+1)/2=n(n-1)/2$. The determinant takes only the values $\pm 1$ there, so $SO(n)$ is an open and closed part of $O(n)$ and has the same manifold dimension:It is closed in matrix space and bounded because every orthogonal matrix column has norm one. The Heine-Borel theorem proves compactness. Finally the tangent space is the derivative kernel,Thus its translated tangent space is the vector space of skew-symmetric matrices.

Write $Ph=\mathbb{E}h(X)$ and $P_{n}h=n^{-1}{\sum }_{i}h(X_i)$ for the expectation and empirical measure. A sufficient bracketing of a function class condition is that, for every $\epsilon >0$, finitely many brackets $[l_j,u_j]$ cover $\mathcal{H}$, with measurable integrable endpoints and $P(u_j-l_j)<\epsilon$. The uniform strong law from finite L1 bracketing then givesHere the observations are independent and identically distributed. For an uncountable class, one either assumes a measurable supremum, for example through a pointwise separable function class, or formulates the conclusion as a pathwise bound on a common probability-one event.

For the parameterized class, the Heine-Borel theorem makes $\Theta$ compact. Put $M(x)={\sup }_{\theta \in \Theta }|q(\theta ,x)|$. A countable dense subset of $\Theta$ gives the same supremum because of continuity, so $M$ is measurable. Define the modulusThe supremum is measurable by taking a countable dense subset of the compact set of admissible pairs. For each $x$, uniform continuity on $\Theta$ gives $w_{\delta }(x)\to 0$. Also $w_{\delta }\le 2M$, and $PM<\infty$. Thus the dominated convergence theorem gives $P{w}_{\delta }\to 0$.

Choose a finite net of radius $\delta$ ${\theta }_1,\dots ,{\theta }_N$ in $\Theta$. The lower and upper envelopes of $q(\theta ,\cdot )$ over each closed ball $\Theta \cap \overline{B}({\theta }_j,\delta )$ are measurable integrable function brackets; the same countable-dense-set argument applies within each such compact ball. Their widths are at most $w_{2\delta }$. They cover the whole class, so its bracketing of a function class condition follows. Moreover, dominated convergence shows that $\theta \mapsto Pq(\theta ,\cdot )$ is continuous. Both empirical and population functions therefore have a supremum over a common countable dense parameter set. Applying the uniform strong law from finite L1 bracketing proves uniform almost-sure convergence over the entire compact parameter set, rather than merely convergence at each fixed parameter.

For the exponential family,where $A={\sup }_{\Theta }|\theta |$ and $B={\sup }_{\Theta }|K(\theta )|$. Thus a weak sufficient condition is bounded $K$ on $\Theta$, $f_0(X)>0$ almost surely, and integrability of $|X|+|\log f_0(X)|$ under the sampling law. No positive lower bound on $f_0$ over the whole real line is required. Zeros off the sampling support are harmless: choose arbitrary finite versions of the log-densities on that common null set when applying the function-class theorem.

If the sampling law is $f({\theta }_0,\cdot )$, one convenient assumption is that $\Theta$ is a compact subset of the interior of the finite domain of the cumulant function of an exponential family, andOn that interior, $K$ is continuous, hence bounded on $\Theta$. Finiteness of $K$ at ${\theta }_0\pm a$ for some $a>0$ implies ${\mathbb{E}}_{{\theta }_0}{e}^{a|X|}<\infty$, hence ${\mathbb{E}}_{{\theta }_0}|X|<\infty$. The displayed integral gives the remaining integrability. If the observations come from an arbitrary unrelated law, assumptions on $K$ and $f_0$ alone cannot control that law's tails; the sampling-law integrability must then be stated explicitly.

If $H$ is a finite-dimensional Hilbert space, a choice of orthonormal basis identifies its closed unit ball with a closed bounded ball in a finite-dimensional Euclidean space. The Heine-Borel theorem makes it compact in the norm topology.

Conversely, if $H$ is infinite-dimensional, choose a unit vector $e_1$ and inductively choose $e_{n+1}$ orthogonal to the span of $e_1,\dots ,e_n$. The finite span is a proper closed subspace, so its orthogonal complement contains a nonzero vector that can be normalized. The resulting orthonormal sequence lies in the closed unit ball and satisfiesIt has no norm-convergent subsequence. Sequential compactness is equivalent to compactness in a metric space, hence the closed ball is not compact. This proves the norm-compact unit ball criterion: a Hilbert space has a norm-compact closed unit ball exactly when it is finite-dimensional.

The sphere is the inverse image of $\{1\}$ under the continuous map $u\mapsto \Vert u{\Vert }^2$, so it is closed in ${\mathbb{R}}^3$. It is bounded. The Heine-Borel theorem therefore makes it compact. Every metric space is Hausdorff: distinct points have disjoint open balls of radii smaller than half their distance. The Euclidean subspace metric supplies that property here. Hence $S^2$ is compact and Hausdorff.

Throughout this question the Banach spaces and Hilbert spaces are complex. Separability is an additional standing hypothesis, not part of the general definition of a Hilbert space. Here and below unit ball means the closed unit ball. An open unit ball in positive finite dimension is not compact. In finite dimension, equivalence of norms identifies the closed unit ball with a closed bounded subset of Euclidean space, so the Heine-Borel theorem gives compactness in the norm topology.

In an infinite-dimensional Hilbert space, the Gram-Schmidt process constructs an infinite orthonormal sequence $(e_n)$. Put $f_n=e_n/\sqrt{2}$. Then $\Vert f_n\Vert =1/\sqrt{2}\le 1$ and, for $m\ne n$, $\Vert f_n-f_m{\Vert }^2=(1+1)/2=1$. No subsequence is a Cauchy sequence, contradicting sequential compactness. Thus the closed unit ball is norm compact exactly in finite dimension.

The finite-dimensional implication again follows from equivalence of norms and the Heine-Borel theorem. In an infinite-dimensional Banach space, inductively apply Riesz lemma to the proper closed finite-dimensional linear span $Y_n=\mathrm{span}(f_1,\dots ,f_n)$ to obtain a unit vector $f_{n+1}$ with $\mathrm{dist}(f_{n+1},Y_n)>1/2$. Therefore $\Vert f_n-f_m\Vert >1/2$ for distinct indices, precluding a Cauchy subsequence.

For completeness, Riesz lemma here follows by taking $z\notin Y$, setting $d=\mathrm{dist}(z,Y)>0$, choosing $y\in Y$ with $\Vert z-y\Vert <2d$, and putting $f=(z-y)/\Vert z-y\Vert$. Its distance from $Y$ exceeds $1/2$. This proves the closed unit ball is norm compact if and only if the Banach space is finite-dimensional. The printed hint's equality of all pairwise distances is unnecessary; the separation inequality is what the argument supplies.

The closure is closed and bounded in ${\mathbb{R}}^2$, hence compact by the Heine-Borel theorem. It is nonempty. The extreme value theorem therefore gives a point of the closure at which the continuous function attains its maximum. The closure, rather than the open domain, is essential.

Let $K$ be a compact set of a Hausdorff space $X$. For any $x\notin K$ and each $y\in K$, the Hausdorff property gives disjoint open sets $U_y\ni y$ and $V_y\ni x$. The sets $U_y$ cover $K$; compactness supplies a finite subcover $U_{y_1},\dots ,U_{y_r}$. The open set $V={\bigcap }_i{V}_{y_i}$ contains $x$ and misses every $U_{y_i}$, hence misses $K$. Thus $X\setminus K$ is open, and every compact subset of a Hausdorff space is closed. The empty-set case is immediate.

Since $\mathbb{R}$ with its usual topology is Hausdorff, its compact sets are closed. The open cover $\{(-n,n):n\ge 1\}$ of $\mathbb{R}$ has a finite subcover of any compact set $K$. Taking the largest index in that subcover gives $K\subset (-N,N)$, so $K$ is a bounded set. This proves the required direction of the Heine-Borel theorem without assuming its converse.

If a nonempty subset $S\subset \mathbb{R}$ is not an interval, there are $a,b\in S$ and $c\notin S$ with $a<c<b$. The disjoint nonempty relatively open sets $S\cap (-\infty ,c)$ and $S\cap (c,\infty )$ form a separation, so $S$ is not connected.

Conversely, suppose an interval $I$ had a separation into relatively open nonempty sets $A,B$. Choose $a\in A,b\in B$, interchanging the names if necessary so $a<b$. Put $c=\sup (A\cap [a,b])$. Because $B$ contains a relative neighbourhood of $b$, $c<b$; because $A$ contains a relative neighbourhood of $a$, $c>a$. Also $c\in [a,b]\subset I$. If $c\in A$, relative openness supplies points of $A$ just above $c$, contradicting the supremum. If $c\in B$, a relative neighbourhood of $c$ excludes points of $A$ approaching $c$ from below, again contradicting the supremum. Thus every interval is connected, including a singleton and unbounded or open intervals.

TakeIt is closed, because its only accumulation point is $0$ and that point is included, and bounded. The Heine-Borel theorem makes it compact. The connected components of its complement are preciselyEach is an interval and therefore connected; none can be enlarged inside the complement across one of the intervening points of $K$.

The closed bounded set $A\subset {\mathbb{R}}^m$ is compact by the Heine-Borel theorem. Choose a convergent subsequence $x_{n_j}\to x_0$. For fixed $N$ and sufficiently large $j$, monotonicity givesContinuity and passage to the limit give $g_N(x_0)\ge \delta$ for every $N$, hence ${\lim }_N{g}_N(x_0)\ge \delta$.

For the deduction, set $g_n=f_n-f$. These functions are continuous, decrease pointwise, and tend to zero. If convergence were not uniform, some $\delta >0$ would admit $x_n$ with $g_n(x_n)\ge \delta$. The result just proved would produce $x_0$ with $\lim g_n(x_0)\ge \delta$, contradicting pointwise convergence. Therefore $f_n\to f$ uniformly. This is Dini's theorem.

The sum is nonempty because $A$ and $B$ are nonempty. It is a bounded set: if $|a|\le M_A$ and $|b|\le M_B$, then $|a+b|\le M_A+M_B$.

To prove that it is a closed set, let $x_j=a_j+b_j\in A+B$ be a convergent sequence with $x_j\to x$. By the Heine-Borel theorem, $A$ is a compact set, so $(a_j)$ has a convergent subsequence $a_{j_r}\to a\in A$. ThenSince $B$ is closed, $x-a\in B$, and hence $x=a+(x-a)\in A+B$. Therefore $A+B$ is nonempty, closed and bounded, so $A+B\in \mathcal{K}$.