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In a metric space $(X,d)$, an open ball of radius $r>0$ about $a$ is $B_d(a,r)=\{x\in X:d(a,x)<r\}$. A closed ball instead uses $d(a,x)\le r$. The strict inequality matters: a point at distance exactly $r$ belongs to the closed ball and is excluded from the open ball. In particular, $(1,0)$ is excluded from all three open unit balls above, and included in all three closed unit balls.

The three closed unit balls centred at zero are $|x_1|+|x_2|\le 1$, $x_1^2+x_2^2\le 1$, and $\max \{|x_1|,|x_2|\}\le 1$. Their shapes are a diamond (a rotated square), a disk with a circle as its boundary, and an axis-aligned square. A ball need not look round! The corresponding open balls are the interiors of these shapes. At $(0.8,0.8)$, Manhattan distance from zero is $1.6$ and Euclidean distance is $\sqrt{1.28}\approx 1.131$, whereas Chebyshev distance is $0.8$: only the square's open unit ball contains this point.

These shapes illustrate $B_1(0,1)\subseteq B_2(0,1)\subseteq B_{\infty }(0,1)$, with either the open ball or closed ball convention used consistently. Their outlines are included in the displayed closed balls. In an arbitrary metric space, however, points at distance exactly $r$ need not form the topological boundary; the discrete metric below shows why this distinction matters.

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 Cayley graph turns multiplication in a group into travel: the chosen generators of a group are the allowed one-step moves, and a word over an alphabet becomes a route. This lets us study a finitely generated group using graph distances, closed balls, and geodesics, rather than only equations between products.

Let $G$ be a group with a finite generating set of a group $S$ satisfying $S=S^{-1}$ and $1\notin S$. Such a symmetric subset of a group includes the inverse element of every allowed move. The Cayley graph ${\mathrm{Cay}}_S(G)$ has vertex set $G$ and joins $g$ to $gs$ for each $s\in S$. We choose right multiplication for these moves. Reversing the move uses $s^{-1}$, so for graph distance we regard $g\to gs$ and $gs\to g$ as the two orientations of one undirected edge. An involution therefore gives one edge, not two parallel edges. Each edge has length one, regardless of its length in a drawing.^[ref]

A sequence $s_1,\dots ,s_n$ labels the walkIt ends at $h$ exactly when $g^{-1}h=s_1\cdots s_n$. Thus the Cayley graph is a connected graph precisely because $S$ generates $G$, and its graph distance on vertices is the word metric. Left multiplication by $a$ sends the edge from $g$ to $gs$ to the edge from $ag$ to $(ag)s$, preserving its generator label: this is a group action by graph automorphisms and isometries. Viewing groups as symmetries of geometric objects is the motivation of Clay and Margalit's introduction.^[ref]

The useful geometric object includes its allowed moves, not just its drawing. Cayley graphs of the integers show how changing the moves adds shortcuts; word metric balls in the integer lattice show how the same group can acquire visibly different closed balls. Meier's book develops this passage from group operations to graphs as its central motivation.^[ref]

The discrete metric sets $d(x,x)=0$ and $d(x,y)=1$ for distinct points. It induces the discrete topology, and its Cauchy sequences are exactly the eventually constant sequences.

Every nonempty set equipped with the discrete metric is a bounded set, even if it is an infinite set. For $0<r<1$, both the open ball and closed ball about $a$ are the singleton $\{a\}$. At $r=1$, the open ball is still $\{a\}$, but the closed ball is the whole space; for $r>1$, both are the whole space. Every subset is both an open set and a closed set. Consequently the boundary of the radius-one closed ball is empty, even though every point other than $a$ is at distance exactly one. Also, the closure of $B_d(a,1)$ is $\{a\}$, rather than the radius-one closed ball.

Distribute a positive electric charge $Q$ uniformly through a closed ball of radius $R$ in vacuum, with no external field. Its volume charge density is $3Q/(4\pi R^3)$. For $0<r<R$, the enclosed fraction is $(r/R{)}^3$. Applying Gauss's law on a concentric Gaussian surface givesThe electric field grows linearly inside and decays as an inverse square outside. With $E_R=Q/(4\pi {\epsilon }_0{R}^2)$, its values at $r/R=1/2,1,2$ are $E_R/2,E_R,E_R/4$. The field vanishes at the centre by spherical symmetry and is continuous at the surface because this ideal volume distribution has no singular surface layer. Its interior behavior differs from a conductor in electrostatic equilibrium, where mobile charge redistributes.

Starting with an arbitrary point in a finite metric space, repeatedly select a point maximizing its distance to the selected set. Maintain each point's distance to its nearest selected center; each new center updates all those distances, giving $O(k|V|)$ distance evaluations for $k$ centers.

For the metric k-center problem, let $R$ be the final covering radius and add a point realizing $R$ to the $k$ selected centers. Every pair among these $k+1$ points has distance at least $R$: at each selection the current maximum distance is at least its final value. If the optimum radius is $r^{\ast }$, two of these points belong to the same optimal radius-$r^{\ast }$ closed ball, so the triangle inequality gives $R\le 2r^{\ast }$. The case $R=0$ is already optimal. This proves a factor-two approximation algorithm.

Under the Picard-Lindelöf theorem's hypotheses on $U\subset \mathbb{R}\times {\mathbb{R}}^d$, extend a solution of a differential equation as far as possible, to a maximal interval $(T_{-},T_{+})$. If $T_{+}<\infty$, the time-state trajectory $(t,y(t))$ must eventually leave every compact subset of $U$. Equivalently, confinement near $T_{+}$ to one compact subset of $U$ permits continuation past $T_{+}$. See Teschl, Section 2.6, Lemma 2.14 and Corollaries 2.15–2.16, pp. 52–53.

The domain restriction matters even for a bounded trajectory. Restrict $y^{\prime }=1$, $y(0)=0$, to $U=\mathbb{R}\times (-1,1)$. Then $y(t)=t$ exists in this domain only for $-1<t<1$; at $t=1/2,3/4,9/10$ its values are $1/2,3/4,9/10$. As $t\uparrow 1$, it approaches the excluded boundary point $(1,1)$ while remaining bounded. Its trajectory cannot fit into any compact subset of $U$ near that endpoint. This is escape from a finite domain, without finite-time blowup in norm. Enlarging the domain to $\mathbb{R}\times \mathbb{R}$ gives the same formula $y(t)=t$ for every real $t$.

Conversely, $y^{\prime }=y$, $y(0)=1$, stays in the compact set $[0,T]\times [1,e^T]$ for every fixed $T>0$. Its growing exponential function is unbounded over the entire future, but has no finite-time blowup. At $T=2$, its largest value on $[0,2]$ is only $e^2\approx 7.38905610$. The relevant question is boundedness before each finite endpoint, rather than one bound valid forever. The tangent function in the diagram under the Picard-Lindelöf theorem illustrates the other possibility: its norm escapes every bound before $T=\pi /2$.

This criterion is useful because an estimate can secure continuation without an explicit formula. In a finite-dimensional vector space, a bounded closed ball is compact; a bounded state trajectory therefore supplies the needed compact set when its time-state closure is inside the domain. That final proviso prevents mistaking a domain boundary for finite-time blowup.

A common sufficient estimate comes from global Lipschitz continuity in the state. Assume $U=I\times {\mathbb{R}}^d$ for an open time interval $I$, with $f$ continuous and a Lipschitz constant $L\ge 0$ valid for all states on each compact time interval under consideration. For $[t_0,T]\subset I$, let $B={\sup }_{t_0\le s\le T}\Vert f(s,0)\Vert$, finite by continuity. Then $\Vert f(t,y)\Vert \le B+L\Vert y\Vert$, and the Gronwall inequality gives, for $t_0\le t\le T$ wherever the solution already exists,For instance, $y^{\prime }=y+1$, $y(0)=0$, has $L=B=1$ and the exact solution $y(t)=e^t-1$, attaining this estimate; at $t=1$ the value is $e-1\approx 1.71828183$. Finite bounds of this kind give continuation throughout $I$, with the analogous estimate backwards in time. When $I=\mathbb{R}$, this gives existence for all real times. If $I$ has a finite endpoint, it gives no continuation beyond that prescribed domain. The underlying linear-growth criterion is ^[ref].

A brief explanation of the continuation mechanism suffices: within the protecting compact set, the continuous function $f$ is bounded, say by $M$. The integral equation yields $\Vert y(t)-y(s)\Vert \le M|t-s|$ near the finite endpoint, so $y(t)$ has a limit $y_{\ast }$. The protecting compact set puts $(T_{+},y_{\ast })$ inside $U$. Restarting the initial value problem there and using local uniqueness joins the trajectories.

The inverse function theorem answers a practical question: can nearby outputs be used to recover nearby inputs uniquely and smoothly? Its useful surprise is that a nonlinear function can be tested using its linear derivative. If the first-order changes preserve all input directions, then, on a sufficiently small region, the nonlinear map can also be reversed. This is the basis for treating a nonlinear map as a coordinate transformation.

Precisely, let $f:U\to {\mathbb{R}}^n$ be continuously differentiable on an open subset $U$ of Euclidean space, let $a\in U$, and put $b=f(a)$. If $Df(a)$ is an invertible linear map, there are open neighbourhoods $U_0$ of $a$ and $V_0$ of $b$ such that $f:U_0\to V_0$ is a bijection with a continuously differentiable inverse function $g$. Thus this restriction is a diffeomorphism. In coordinates the test is $\det Df(a)\ne 0$, using the Jacobian determinant. The derivative of a local inverse map is obtained by inverting the Jacobian matrix, at the corresponding input. ^[ref]

For a one-dimensional example, $f(x)=x^2$ has $f^{\prime }(1)=2$. Near $x=1$, the inverse function is $g(y)=\sqrt{y}$, and the inverse function rule gives $g^{\prime }(1)=1/2$. But $x^2$ has no inverse function on the whole real line, since $x$ and $-x$ give the same output. The theorem supplies a local restriction, not global injectivity. At zero, the derivative test fails; the nonnegative branch still has a square root inverse, but that inverse has no finite derivative at zero. ^[ref]

Here is a two-dimensional example with an invertible derivative everywhere and infinitely repeated outputs:The exponential function turns $s$ into a positive radius, while $\theta$ is the angle in polar coordinates. Restricting, for example, to $|s|<0.35$ and $|\theta |<0.6$ produces the reversible patch pictured below. On its image, with Cartesian coordinates $(u,v)$ and $u>0$, the inverse function isThe arctangent selects the angle near zero. Without an angle restriction, $f(s,\theta +2\pi k)=f(s,\theta )$ for every integer $k$, so there is no single global inverse function. This map is a local diffeomorphism but is not injective.

The short proof idea is to correct a guessed input using the inverse linear response. With $A=Df(a)$, solving $f(x)=y$ is equivalent to finding a fixed point ofAt $a$, the derivative $D{T}_y(a)=I-A^{-1}Df(a)$ vanishes. Continuity makes $T_y$ a contraction mapping on a sufficiently small closed ball, which it maps into itself when $y$ is sufficiently close to $b$. The contraction mapping theorem then gives existence and uniqueness there. Establishing the smooth dependence on $y$ completes the inverse theorem. This explains why the nonlinearity can be tolerated: its correction has a small local slope.^[ref]

The metric k-center problem chooses at most $k$ centers from a finite metric space to minimize the largest distance from a point to its nearest center. For $1\le k\le |V|$, exactly $k$ centers gives the same optimum: extra centers never increase the radius. The restriction that centers belong to $V$ matters; allowing centers anywhere in an ambient metric space is a different problem.

The basic packing argument is short. If a radius-$r$ solution uses $k$ centers, its $k$ closed balls cover $V$. Two points assigned to the same center have distance at most $2r$ by the triangle inequality. Thus $k+1$ points with pairwise distances greater than $2r$ certify that no radius-$r$ solution exists. Greedily selecting a new center only outside all existing radius-$2r$ closed balls therefore uses at most $k$ centers whenever a radius-$r$ solution exists, and its final covering radius is at most $2r$. The farthest-first traversal achieves the same approximation ratio without knowing $r$.

For a nonempty finite set $\{p_1,\dots ,p_m\}$ in finite-dimensional hyperbolic space, the continuous function $R(x)={\max }_{j}d(x,p_j)$ attains its minimum. Indeed, a bounded candidate radius confines all possible improving centers to a closed bounded ball about $p_1$. Such a ball is compact in the Poincare ball model, since its points remain away from the ideal boundary. Minimizing $R$ on it supplies a closed ball of smallest enclosing radius. More generally this argument works for any nonempty bounded set in a proper metric space when the supremum-distance function is finite.

Use closed balls with centres belonging to the space being covered. The covering number iswith value $\infty$ if there is no finite metric net, and value zero for the empty space.

With this intrinsic-centre convention, the proposed monotonicity under subsets is false. Give $R=\{-1,0,1\}$ the usual distance, take $S=\{-1,1\}$, and set $\epsilon =3/2$. A ball centred at $0$ covers $R$, so $N(3/2;R,\rho )=1$. The two points of $S$ are distance $2$ apart, so a ball centred in $S$ contains only its centre, and $N(3/2;S,\rho )=2$. These values are also valid with open balls.

The obstruction is that the useful centre $0$ disappears on passing to the subspace. If centres were allowed anywhere in a fixed ambient space, monotonicity would hold. With intrinsic centres, one instead hasIndeed, discard covering balls missing $S$ and choose one point of $S$ from each remaining ball. The triangle inequality shows that every point of its intersection with $S$ is within $2\epsilon$ of the chosen point. This proves the replacement bound whenever the right side is finite; otherwise it is automatic.

The optimum radius is a distance from some city to a selected city, so it belongs toSort the distinct values of $D$, try them in increasing order, and run part (i)'s greedy procedure with each trial value $r$. Declare a trial successful when all cities are covered using at most $k$ centers; stop a failed trial as soon as a $(k+1)$st center would be selected. The trial $r=r^{\ast }$ must succeed by part (i), so the first successful value $\hat{r}$ satisfies $\hat{r}\le r^{\ast }$. Its output hasThere are $O(n^2)$ candidate distances and $O(n^2)$ work per trial, giving a polynomial-time algorithm with running time $O(n^4)$, plus sorting. Success at $r$ need not mean that radius $r$ itself is feasible; only the radius-$2r$ guarantee is used.

A faster alternative is the farthest-first traversal. Choose any first center, then repeatedly choose a city farthest from its nearest current center until $k$ have been chosen. Let $R$ be the resulting radius. If $R>0$, adjoin a city attaining that distance. At every earlier selection, the maximum distance to the then-current center set was at least $R$, so all pairs among these $k+1$ cities are at distance at least $R$. Two belong to the same optimal radius-$r^{\ast }$ closed ball, giving $R\le 2r^{\ast }$ by the triangle inequality. If $R=0$, the guarantee is immediate. Maintaining nearest-center distances gives $O(nk)$ distance evaluations without knowing $r^{\ast }$.

Neither assertion holds for a general metric space. Take $X=\{0,1\}$ with $d(0,1)=1$ and center $0$. This is a discrete topology: both singleton sets are open and closed. The open ball is $U=\{0\}$, whereas the closed ball is $K=X$. Thus the interior of $K$ is $X$, strictly larger than $U$, and the closure of $U$ is $\{0\}$, strictly smaller than $K$. In general $U\subseteq \mathrm{int}K$ and $\overline{U}\subseteq K$, but the inclusions need not be equalities.

The complete metric space $\mathbb{R}$ supplies a counterexample: $A_n=[n,\infty )$ are descending nonempty closed sets, but no real number belongs to them all.

Now assume shrinking diameters. Choose $x_n\in A_n$. Given $\epsilon >0$, choose $N$ so that $\mathrm{diam}(A_N)<\epsilon$. Whenever $m,n\ge N$, both chosen points lie in $A_N$, so $d(x_m,x_n)<\epsilon$. Thus $(x_n)$ is a Cauchy sequence. By completeness it converges to some $x\in X$. For each fixed $N$, the tail lies in the closed set $A_N$, so its limit lies there too. Hence $x\in {\bigcap }_n{A}_n$. Any two points of this intersection have distance at most every $\mathrm{diam}(A_n)$, so their distance is zero. The shrinking intersection contains exactly one point. This is the Cantor intersection theorem.

For the dense-open conclusion, begin inside an arbitrary nonempty open set $O\subseteq X$; taking $O=X$ gives the requested existence. Since $U_1$ is a dense subset, choose $x_1\in O\cap U_1$ and $0<r_1\le 1/2$ small enough that the closed ball $A_1=\overline{B}(x_1,r_1)$ is contained in $O\cap U_1$. Here $\overline{B}$ denotes the closed ball, not an assumption about the closure of the open ball.

Inductively, the open ball $B(x_n,r_n)$ meets the dense open set $U_{n+1}$. Choose $x_{n+1}$ in that intersection and a positive radius $r_{n+1}\le 2^{-(n+1)}$ so small thatSuch a radius exists by openness and the triangle inequality. These nonempty closed sets descend and have diameters at most $2r_n\to 0$. The preceding result gives a point in every $A_n$, hence in $O$ and every $U_n$. Therefore ${\bigcap }_n{U}_n$ is nonempty; in fact it is dense in $X$. This proves the relevant Baire category theorem directly, without assuming closed balls are compact.

We first prove the stronger compact-square form, in which all the one-variable functions are continuous. There are five fixed pairs of continuous functions $a_q,b_q$ on the unit interval such that every real $f\in C([0,1{]}^2)$ has a representationwith continuous one-variable $G_q$. We construct the inner functions and then prove convergence of the outer functions, rather than assume the Kolmogorov-Arnold representation theorem.

Constructing separating inner functions. Work in the Banach space $\mathcal{X}=C([0,1]{)}^{10}$ with the maximum of the ten supremum norms. For each positive integer $n$, let $U_n$ consist of tuples $(a_q,b_q{)}_{q=1}^5$ for which there are, for each $q$, finite families of closed intervals $I_{qi}$ and $J_{qj}$ of length less than $1/n$, satisfying two conditions. Each point of the unit interval belongs to at least four of the five unions ${\bigcup }_i{I}_{qi}$, and likewise for the $J$ families. Within each fixed $q$, the compact setsare pairwise disjoint for distinct pairs $(i,j)$. Each $K_{qij}$ is a closed interval, by continuity and the intermediate value theorem. These are grid-separated additive coordinates.

The set $U_n$ is open: for a tuple and its finitely many witnessing intervals, the distinct compact image intervals have a positive minimum separation. A sufficiently small uniform change in the inner functions preserves that separation, while the interval-cover conditions do not change.

To prove density, start with any ten continuous functions and any positive approximation tolerance. Choose a common fine partition so that each original function oscillates by less than a small fraction of the tolerance on a partition cell and its immediate neighbours; choose its mesh smaller than $1/(2n)$. Near every internal division point put five small, pairwise disjoint open gaps, one for each $q$. Use these gaps to divide the interval into closed towns for family $q$, including the two endpoints in towns. Gaps from different families are disjoint, so any point misses at most one family; every town has length less than $1/n$ if the gaps are sufficiently small. Do this for both coordinates.

Approximate $a_q$ and $b_q$ by functions which are constant on their respective towns, and interpolate linearly across the gaps. The towns lie near single partition cells, so uniform continuity makes these approximations as close to the original functions as desired. For each $q$, perturb the finitely many town values ${\alpha }_{qi},{\beta }_{qj}$ arbitrarily slightly so that all sums ${\alpha }_{qi}+{\beta }_{qj}$ are distinct. This is possible because each unwanted equality between distinct pairs is a proper affine hyperplane in the finite-dimensional vector space of values; finitely many such hyperplanes cannot fill any open ball. Interpolating the perturbed values still approximates the original functions, and the rectangle images are now distinct singleton sets. Thus $U_n$ is dense.

By the Baire category theorem, choose one tuple in ${\bigcap }_{n\ge 1}{U}_n$. To recall why completeness suffices here, start inside any open ball and successively choose closed balls of positive radii tending to zero, each contained in the preceding ball and in the next dense open $U_n$. Their centers form a Cauchy sequence, and its limit belongs to every ball, hence every $U_n$. Fix this tuple once and for all: it does not depend on $f$.

A quantitative approximation step. Let a continuous residual $r$ have supremum norm $M>0$. Choose $n$ large enough that the oscillation of $r$ on any rectangle of side lengths below $1/n$ is at most $M/20$. For each rectangle $I_{qi}\times J_{qj}$ choose a sample point $(x_{qi},y_{qj})$. Define a one-variable $h_q$ to have constant value $r(x_{qi},y_{qj})/3$ on $K_{qij}$. The image intervals are disjoint, so interpolate linearly in their gaps and extend constantly beyond the outermost ones. Then $h_q$ is continuous on the real line andAt a given $(x,y)$, at most one family fails to contain $x$ in an $I$ town and at most one fails to contain $y$ in a $J$ town. Thus the point belongs to rectangles in $k$ families, where $3\le k\le 5$. For each of these families the summand differs from $r(x,y)/3$ by at most $M/60$. Each remaining summand has absolute value at most $M/3$. ConsequentlyThis proves a strict contraction uniformly over the square, including points in gaps.

Summing the corrections. Set $r_0=f$ and apply that step repeatedly, subtracting the five summands at each stage. The residuals satisfy $\Vert r_m{\Vert }_{\infty }\le (3/4{)}^m\Vert f{\Vert }_{\infty }$, and the corresponding corrections satisfy $\Vert h_{qm}{\Vert }_{\infty }\le (3/4{)}^m\Vert f{\Vert }_{\infty }/3$. If a residual is zero, take all subsequent corrections to be zero. The seriesconverges uniformly on the real line to a continuous function. The residuals tend uniformly to zero, proving the boxed representation. An affine change of each coordinate treats any compact rectangle.

Removing a restriction to a compact domain. For a continuous real function $f$ on the whole plane, setChoose a positive continuous one-variable $A$ by linear interpolation of $A(n)=(n+2)(1+M(n+1))$ at nonnegative integers. Since these values increase, for $n\le t\le n+1$ we have $A(t)\ge (n+2)(1+M(n+1))\ge (1+t)(1+M(t))$. ThereforeLet $\sigma (x)=\frac{1}{2}+{\pi }^{-1}\arctan x$. Transport $b$ to the open square by $\tilde{b}(s,t)=b(\tan (\pi (s-1/2)),\tan (\pi (t-1/2)))$, and set it to zero on the square's boundary. The displayed decay estimate proves continuity at every boundary point, including the corners. Apply the compact-square representation to $\tilde{b}$ and substitute $s=\sigma (x)$, $t=\sigma (y)$ to obtain $b$ using continuous one-variable functions and addition.

Finally multiply this expression by $A(x^2+y^2)$. Multiplication itself uses only addition and one-variable functions, becauseSquaring, negation and scaling are one-variable operations, as are $A$ and $\sigma$. Thus the whole-plane reduction for continuous superposition gives a finite expression of the required kind for every continuous $f$ on the plane, without a boundedness assumption. For a complex-valued $f$, apply the argument to its real and imaginary parts and combine them using addition and the one-variable map $z\mapsto iz$. All the constituent one-variable functions can be taken continuous.

The Baire category theorem states that in a complete metric space, a countable intersection of open dense sets is dense. Equivalently, a nonempty complete metric space cannot be a countable union of nowhere dense sets, where a set is nowhere dense if its closure has empty interior.

Here is the proof. Let $G_1,G_2,\dots$ be open dense sets and let $U$ be any nonempty open set. Choose a point $a_1\in U\cap G_1$ and a radius $0<r_1<1/2$ such that the closed ball $B_1=\overline{B}(a_1,r_1)$ lies inside $U\cap G_1$. Inductively choose $a_{m+1}$ in the nonempty open set $B(a_m,r_m)\cap G_{m+1}$, and then a radius $0<r_{m+1}<2^{-m-1}$ such thatThus the closed balls are nested and their radii tend to zero. For $j,k\ge m$, both centers lie in $B_m$, so $d(a_j,a_k)\le 2r_m$. They form a Cauchy sequence. Completeness gives a limit $a$, and closedness puts it in every $B_m$. Consequently $a\in U\cap {\bigcap }_m{G}_m$. As $U$ was arbitrary, the intersection is dense.

To obtain the equivalent formulation, take complements of the closures of the nowhere dense sets; those complements are open dense. Conversely, complements of open dense sets are closed with empty interior. In particular, a countable closed cover of a nonempty complete metric space must contain a set with nonempty interior.

A subset is nowhere dense when its closure has empty interior. It is of first category, or meagre, when it is a countable union of nowhere-dense subsets; otherwise it is of second category.

The Baire category theorem says that in a complete metric space every countable intersection of open dense subsets is dense. To prove it, start inside an arbitrary nonempty open set and choose a closed ball of positive radius there and in the first dense open set. Inductively choose a closed ball inside the interior of the preceding ball and the next dense open set, with radius less than $2^{-n}$. The centres are Cauchy. Completeness gives a limit lying in all the nested closed balls, hence in all the dense open sets and the original open set. Thus a nonempty complete metric space is not meagre in itself.

For $p<r\le \infty$, put $E_m=\{x\in {\ell }^p:\Vert x{\Vert }_p\le m\}$, viewed inside ${\ell }^r$. Necessarily $p<\infty$. Each $E_m$ is closed: convergence in the ${\ell }^r$ norm gives coordinatewise convergence, and every finite partial sum of $\sum |x_i{|}^p$ is bounded by $m^p$ in the limit. Taking the supremum of partial sums proves the assertion.

It has empty interior. Given a point in it and any small ${\ell }^r$ radius, perturb a block of $N$ coordinates by entries of modulus $\delta N^{-1/r}$, choosing signs to avoid cancelling the existing coordinates. The ${\ell }^r$ distance is $\delta$, while the resulting ${\ell }^p$ norm is at least $\delta N^{1/p-1/r}$, exceeding $m$ for large $N$. For $r=\infty$, use entries of modulus $\delta$ on that block. Hence each $E_m$ is nowhere dense, andFor $r=p$, including $p=\infty$, the space is Banach and so second category in itself by Baire. There is no case $r>p$ when $p=\infty$.

Let $p,q$ lie in the closed ball of center $z$ and radius $R$. On their unique geodesic, the Hadamard squared-distance inequality givesThus every point of that geodesic lies in the ball. By uniqueness there are no other geodesics to consider, so every closed ball is geodesically convex.

The Lusternik-Schnirelmann-Borsuk theorem has the following two equivalent covering formulations. For every integer $d\ge 0$, a cover of the sphere $S^d$ by $d+1$ closed sets has a member containing an antipodal pair. The same assertion holds with open sets in place of closed sets. Thus $d+1$ antipodal-pair-free open sets, or $d+1$ antipodal-pair-free closed sets, cannot cover $S^d$. The dimension-zero case simply says that one set covering the two-point sphere contains both points.

Here is why the two versions agree. A finite open cover of a compact metric space admits a closed shrinking that still covers: sufficiently small closed balls subordinate to the open cover can be grouped according to their containing open member. Applying the closed version to that shrinking proves the open version. Conversely, if a nonempty closed subset $C$ of $S^d$ avoids antipodal pairs, compactness gives positive distance between $C$ and $-C$. A sufficiently small open neighbourhood of $C$ still avoids antipodal pairs. Enlarge each member of a hypothetical closed counterexample in this way; the open version rules it out. Empty members cause no difficulty.

Another common equivalent formulation is the Borsuk-Ulam theorem: every continuous $f:S^d\to {\mathbb{R}}^d$ has $f(x)=f(-x)$ for some $x$, or, equivalently, every continuous odd function $g:S^d\to {\mathbb{R}}^d$ has a zero. For example, if $d+1$ antipodal-pair-free open sets covered $S^d$, a subordinate partition of unity $({\varphi }_1,\dots ,{\varphi }_{d+1})$ would give the odd functionIt cannot vanish, since some ${\varphi }_i(x)>0$, whereas antipodal-pair-freeness forces ${\varphi }_i(-x)=0$. In the other direction, if an odd function $g:S^d\to {\mathbb{R}}^d$ never vanishes, choose $d+1$ vectors $u_i$ forming a regular simplex centred at the origin in ${\mathbb{R}}^d$. For $d\ge 1$, the open sets $\{x:u_i\cdot g(x)>0\}$ cover $S^d$ and none contains an antipodal pair, contradicting the covering theorem.

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set, $d<\infty$, and $f:U\to {\mathbb{R}}^d$ a continuous function. Through every $(t_0,y_0)\in U$ there is a local solution of a differential equation: a continuously differentiable function $y:I\to {\mathbb{R}}^d$ on an open interval containing $t_0$, satisfying the displayed initial value problem pointwise. No Lipschitz condition is assumed.^[ref]

Continuity of a proposed rate law therefore guarantees at least one trajectory for a short time. For a physical model, this is a first consistency check: can the stated rule and initial condition actually be followed? Prediction also needs uniqueness. The delayed solutions of a square-root differential equation below show how the same rule and initial condition can leave a departure time undetermined. The Picard-Lindelöf theorem supplies a sufficient additional hypothesis for uniqueness; failure of that hypothesis alone does not prove nonuniqueness.

For the initial value problem $y^{\prime }=2\sqrt{|y|}$, $y(0)=0$, a forward solution of a differential equation can stay zero until any $c\ge 0$, then follow $y=(t-c{)}^2$. The delayed solutions of a square-root differential equation verify this freedom exactly.

Local existence gives neither a global solution of a differential equation nor a bound on all future values. The continuous function $f(t,y)=y^2$ already permits finite-time blowup; the finite-dimensional continuation criterion explains the additional issue of extending a trajectory within its domain.

Proof outline: choose a compact set $K=[t_0-a,t_0+a]\times {\overline{B}}_r(y_0)\subset U$. By the extreme value theorem, $M={\max }_K\Vert f\Vert <\infty$. Take $0<h\le a$ with $Mh\le r$. Forward Euler method polygons remain in the closed ball and satisfy $\Vert y_n(t)-y_n(s)\Vert \le M|t-s|$. The Arzelà-Ascoli theorem yields a convergent subsequence with uniform convergence. Uniform continuity of $f$ on $K$ makes their integral equation defect tend uniformly to zero, so the limit satisfiesThe fundamental theorem of calculus gives a continuously differentiable function. Repeat backward and join at $t_0$, where both derivatives equal $f(t_0,y_0)$.^[ref]

The compact set here uses a finite-dimensional vector space; bounded closed balls in an arbitrary Banach space need not be compact. When $M=0$, the constant solution of a differential equation $y=y_0$ works on the entire time interval in $K$. In general the proof selects a subsequence, without asserting that all polygons converge to one unique trajectory.