The Brouwer fixed-point theorem

Theorem

The Brouwer fixed-point theorem

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Statement

Let \( n \ge 1 \) and let \( D^n = \{ x \in \mathbb{R}^n : |x| \le 1 \} \) be the closed unit ball, with boundary sphere \( S^{n-1} = \{ x \in \mathbb{R}^n : |x| = 1 \} \). If \( f : D^n \to D^n \) is continuous, then there exists \( x_0 \in D^n \) with \( f(x_0) = x_0 \).

Why it matters

Brouwer's theorem is the archetype of an "existence without construction" result: it guarantees a solution to \( f(x) = x \) using only the topological shape of the domain and the continuity of \( f \), with no algebraic or analytic information about \( f \) at all. This makes it a load-bearing lemma across mathematics far outside topology proper — it underlies the existence of Nash equilibria in game theory, the existence of solutions to certain systems of nonlinear equations, fixed-point arguments in the proof of the Perron–Frobenius theorem, and the invariance-of-domain circle of results in topology itself.

It is also the cleanest illustration of a general proof strategy in algebraic topology: convert a geometric/analytic problem ("does \( f \) have a fixed point?") into an algebraic one ("does a certain group homomorphism exist?"), then kill the algebraic problem by a one-line computation. The heavy lifting — building the invariant that distinguishes a disc from its boundary sphere — is done once, by singular homology, and then reused for free.

Hypotheses

Domain is the closed ball \( D^n \) (compact, convex). Drop compactness and work on the open ball \( B^n = \{ |x| \lt 1 \} \) instead: the map \( f(x) = \tfrac{1}{2}(x + v) \) for a fixed unit vector \( v \), restricted to \( B^n \), pushes every point strictly toward the missing boundary point \( v \) and has no fixed point in \( B^n \).

\( f \) is continuous. On \( D^1 = [-1,1] \) let \( f(x) = 1 \) for \( x \lt 0 \) and \( f(x) = -1 \) for \( x \ge 0 \); this discontinuous map sends every point away from itself and has no fixed point.

\( f \) maps \( D^n \) into itself (not into a larger or different space). The identity-like map \( f(x) = 2x \) is continuous on \( D^n \) but sends \( D^n \) outside itself (except at \( 0 \), which happens to be fixed here); in general, allowing the codomain to be a strict superset removes the topological obstruction and fixed points need not exist, e.g. \( f(x) = x + v \) on \( D^n \subset \mathbb{R}^n \) for fixed \( v \neq 0 \), viewed as a map \( D^n \to \mathbb{R}^n \).

The domain is genuinely a topological (or homeomorphic image of a) ball — no "hole". Take the annulus \( A = \{ x \in \mathbb{R}^2 : 1 \le |x| \le 2 \} \) and let \( f \) be rotation by \( 90^\circ \): continuous, self-map, but no fixed point, because \( A \) deformation retracts to a circle rather than a point and carries no analogous vanishing-homology obstruction.

Proof

Suppose, for contradiction, that \( f : D^n \to D^n \) is continuous and \( f(x) \neq x \) for every \( x \in D^n \).

Standard reductio: we derive a contradiction with a known homological fact, forcing a fixed point to exist. A

For each \( x \in D^n \) set \( v(x) = x - f(x) \neq 0 \), and define \( r(x) \) to be the unique point where the ray starting at \( f(x) \) and passing through \( x \) meets \( S^{n-1} \), taking the intersection point reached by continuing forward from \( x \) (i.e. with parameter \( t \ge 0 \) in \( x + t\,v(x) \)).

Well-defined: writing \( r(x) = x + t v(x) \) and imposing \( |r(x)|^2 = 1 \) gives the quadratic \( |v(x)|^2 t^2 + 2(x\cdot v(x))\,t + (|x|^2 - 1) = 0 \). Since \( |x| \le 1 \), the constant term \( |x|^2-1 \le 0 \), so the two roots multiply to a number \( \le 0 \); hence a real root \( t \ge 0 \) exists (and is unique once we insist on the larger root, since the quadratic is nonnegative for large \( |t| \) and non-positive at \( t=0 \)). B

\[ t(x) = \frac{-(x\cdot v(x)) + \sqrt{(x\cdot v(x))^2 - |v(x)|^2\big(|x|^2-1\big)}}{|v(x)|^2}, \qquad r(x) = x + t(x)\,v(x). \]

Explicit closed form for the nonnegative root, using the quadratic formula on Step 2. The discriminant \( (x\cdot v(x))^2 - |v(x)|^2(|x|^2-1) \) is a sum of a square and a nonnegative quantity (since \( |x|^2 - 1 \le 0 \)), hence \( \ge 0\), so the square root is real. C

The map \( x \mapsto v(x) = x - f(x) \) is continuous and nowhere zero on \( D^n \) (by Step 1), so \( |v(x)|^2 \gt 0 \) everywhere; hence \( r : D^n \to \mathbb{R}^n \) given by the formula in Step 3 is a composite of continuous, nowhere-undefined operations (subtraction, dot product, square root of a nonnegative continuous function, division by a nowhere-zero continuous function), and is therefore continuous. Also \( |r(x)| = 1 \) for all \( x \) by construction, so \( r : D^n \to S^{n-1} \).

Continuity of algebraic combinations of continuous functions, plus continuity of \( \sqrt{\cdot} \) on \( [0,\infty) \); no division-by-zero or negative-square-root failure occurs because \( |v(x)| \gt 0\) and the discriminant is \( \ge 0 \) throughout, by Step 3's remark. B

If \( |x| = 1 \) then \( r(x) = x \), i.e. \( r \) restricted to \( S^{n-1} \) is the identity map.

When \( |x|=1\), the constant term \( |x|^2-1 = 0\) in the quadratic of Step 2, so \( t=0 \) is a root; since \( f(x) \in D^n \) and \(D^n\) is convex, the segment from \( f(x) \) to \( x \) lies in \( D^n \) and meets \( S^{n-1} \) only at its endpoint \( x \), so the forward intersection with \( t \ge 0 \) selected in Step 3 is exactly \( t=0 \), giving \( r(x)=x\). B

Thus \( r : D^n \to S^{n-1} \) is a continuous retraction, i.e. \( r \circ \iota = \mathrm{id}_{S^{n-1}} \), where \( \iota : S^{n-1} \hookrightarrow D^n \) is the inclusion.

Immediate from Steps 4 and 5: \(r\) is continuous, \(D^n\)-valued becomes \(S^{n-1}\)-valued, and it fixes \(S^{n-1}\) pointwise, which is precisely the definition of a retraction of \(D^n\) onto \(S^{n-1}\). A

Apply the singular homology functor \( H_{n-1}(-;\mathbb{Z}) \) to \( \iota \) and \( r \): this yields group homomorphisms \( \iota_* : H_{n-1}(S^{n-1}) \to H_{n-1}(D^n) \) and \( r_* : H_{n-1}(D^n) \to H_{n-1}(S^{n-1}) \) with \( r_* \circ \iota_* = \mathrm{id}_{H_{n-1}(S^{n-1})} \).

Functoriality of singular homology: it sends composites of continuous maps to composites of the induced homomorphisms and identities to identities, so \( (r\circ\iota)_* = r_*\iota_* = (\mathrm{id}_{S^{n-1}})_* = \mathrm{id}_{H_{n-1}(S^{n-1})} \). C

\[ H_{n-1}(D^n;\mathbb{Z}) = 0 \quad (\text{since } D^n \text{ is convex, hence contractible}), \qquad H_{n-1}(S^{n-1};\mathbb{Z}) \cong \mathbb{Z} \quad (\text{for } n \ge 1, \text{ reading } H_0(S^0)=\mathbb{Z}^2 \text{ for } n=1). \]

Homotopy invariance of singular homology gives \( H_k(D^n) \cong H_k(\mathrm{pt}) \), which vanishes for \( k \gt 0 \) (standard computation); the homology of spheres, \( H_k(S^{n-1}) \cong \mathbb{Z} \) for \( k = n-1 \ge 1\) and \( 0 \) otherwise, is the standard reduced-homology-of-spheres computation (by induction via Mayer–Vietoris or the long exact sequence of the pair \( (D^n, S^{n-1}) \)); for \( n=1 \), \( S^0 \) is two points and \( \widetilde H_0(S^0) \cong \mathbb{Z} \), which plays the same role. C

A homomorphism \( \mathbb{Z} \xrightarrow{\iota_*} 0 \xrightarrow{r_*} \mathbb{Z} \) cannot compose to the identity on \( \mathbb{Z} \), since the composite factors through the zero group and so is the zero map, whereas \( \mathrm{id}_{\mathbb{Z}} \) is not zero. This contradicts Step 7.

Elementary fact about group homomorphisms: any map through the trivial group is the zero homomorphism, and the zero homomorphism on \( \mathbb{Z} \ne 0 \) is not the identity. A

Hence the assumption in Step 1 is false: there is no continuous fixed-point-free self-map of \( D^n \). Equivalently, every continuous \( f : D^n \to D^n \) has a fixed point.

Reductio ad absurdum discharged: the contradiction obtained in Step 9 refutes the hypothesis of Step 1. A

Result

\( f : D^n \to D^n \text{ continuous} \;\Longrightarrow\; \exists\, x_0 \in D^n,\; f(x_0) = x_0 \)

Reading. You cannot continuously stir, stretch, or fold a disc into itself without leaving at least one point exactly where it started — no matter how wild the map, as long as it is continuous and stays inside the disc, somewhere is pinned in place.

Scope. Holds for the closed ball \( D^n \) in \( \mathbb{R}^n \) for every \( n \ge 1 \), and more generally for any space homeomorphic to \( D^n \) (any compact convex subset of \( \mathbb{R}^n \) with nonempty interior, via a homeomorphism to \( D^n \)), and even for any retract of \( D^n \) (any space \( Y \) admitting continuous \( D^n \xrightarrow{p} Y \xrightarrow{i} D^n \) with \( p\circ i = \mathrm{id}_Y \)) by transporting the fixed point of \( i\circ f\circ p \) back through \( i \). It does not extend as stated to the open ball, to unbounded convex sets, or to infinite-dimensional balls (Kakutani's counterexample in \( \ell^2 \)).

Corollaries & converses

No-retraction theorem is logically equivalent to Brouwer: Step 6 shows Brouwer \(\Rightarrow\) no retraction \(D^n \to S^{n-1}\) fails to exist is what makes the proof go through, and conversely the existence of such a retraction is exactly what a fixed-point-free map would build, so each statement can be deduced from the other in a few lines.
Perron–Frobenius and the existence of a stationary distribution for a finite Markov chain follow by applying Brouwer to the action of a stochastic matrix on the standard simplex (itself homeomorphic to a disc).
Nash equilibrium existence in finite games follows by applying (a multivalued generalisation via) Brouwer/Kakutani to the best-response map on a product of simplices.
Converse fails: the fixed-point property does not characterise discs. Any contractible compact convex set has it, and even some non-convex spaces (e.g. a "warped disc" homeomorphic to \(D^n\)) have it for the trivial reason that they are homeomorphic to \(D^n\); but the property can also hold or fail for spaces with no relation to discs, so "has the fixed point property" is strictly weaker than "is a disc" — it is not an iff.
The theorem gives no information about the number or location of fixed points, only existence of at least one; some maps (e.g. the identity) fix every point, others fix exactly one.

Fails without

Convexity/topology of the domain dropped (annulus instead of disc): rotation by \(90^\circ\) on the annulus \( \{1 \le |x| \le 2\} \subset \mathbb{R}^2 \) is a continuous self-map with no fixed point, because the annulus is not simply connected/contractible and the homological obstruction used in Steps 7–9 is no longer available (\(H_1\) of the annulus is \(\mathbb{Z}\), not \(0\), so the retraction argument does not produce a contradiction).
Continuity dropped: on \(D^1=[-1,1]\), the map \(f(x) = -\mathrm{sgn}(x)\) for \(x \ne 0\), \(f(0)=1\) (any discontinuous convention at \(0\)) sends every point strictly away from itself; no fixed point exists because the intermediate value argument (the \(n=1\) shadow of Step 1–10) relies entirely on continuity.
Boundedness/compactness dropped (open ball): on the open ball \(B^n\), \(f(x) = \tfrac{1}{2}(x+v)\) for a fixed \(v \in S^{n-1}\) maps \(B^n\) into itself continuously (check: \(|f(x)| \le \tfrac12(|x|+1) \lt 1\) whenever \(|x|\lt1\)) but the unique candidate fixed point of the affine map, \(x=v\), lies on the missing boundary, so no fixed point exists in \(B^n\).
Finite dimension dropped: Kakutani's example on the closed unit ball of \(\ell^2\), \(f(x_1,x_2,\dots) = \big(\sqrt{1-\|x\|^2},\, x_1,\, x_2,\dots\big)\), is continuous, ball-to-ball, and fixed-point free, because the ball fails to be compact in infinite dimensions and the homology computation of Step 8 no longer pins down a nontrivial obstruction in the same way.

Common errors

Believing the theorem gives a unique fixed point — it only guarantees at least one; contraction-mapping-style uniqueness needs an extra Lipschitz/contraction hypothesis (Banach fixed-point theorem), which Brouwer does not assume.
Trying to apply the theorem to open discs or open balls, forgetting compactness of the domain is essential (see Fails without).
Confusing "retraction exists" with "deformation retraction exists" — the proof only needs the weaker notion (a continuous \(r\) with \(r\circ\iota=\mathrm{id}\)); no homotopy between \(\mathrm{id}_{D^n}\) and \(\iota\circ r\) is claimed or needed.
Misapplying the theorem to non-self-maps, e.g. \(f:D^n\to\mathbb{R}^n\) with image not contained in \(D^n\) — the codomain-equals-domain hypothesis is essential and silently dropping it invalidates the conclusion.
Thinking the proof is "purely analytic": the construction of the retraction \(r\) (Steps 2–6) is elementary algebra, but the actual obstruction (Steps 7–9) is irreducibly homological — no purely calculus-based argument for general \(n\) is being smuggled in.
Assuming the fixed point can be computed or approximated directly from the proof — the argument is a pure existence proof (via reductio) and gives no algorithm; constructive/computational fixed-point-finding (e.g. Sperner's lemma based, or Scarf's algorithm) is a separate, harder question.

Discussion

Brouwer proved the theorem for \(n=3\) in 1909–1911 using a degree-theoretic argument (closely related to, but predating in rigorous form, the homological machinery used above); the general-\(n\) statement and various proofs (simplicial/combinatorial via Sperner's lemma, homological via singular or simplicial homology, and analytic via Stokes' theorem for smooth maps) were developed over the following decades. The homological proof given here is the modern "textbook" route because it isolates exactly what the theorem needs: a homotopy invariant that distinguishes the sphere from the disc it bounds.

The combinatorial route via Sperner's lemma deserves mention: triangulate \(D^n\) finely, colour vertices according to a Sperner-admissible rule tied to \(f\), and Sperner's lemma guarantees a "fully coloured" simplex in every triangulation; taking finer and finer triangulations and a compactness argument produces a fixed point in the limit. This proof is constructive in spirit (it underlies practical fixed-point-approximation algorithms) even though it still only proves existence in the limit, not an exact finite computation.

The theorem is a genuinely topological fact, not a metric or linear one: it survives any homeomorphism of the domain (any topological disc works, Steps 1–10 go through verbatim after transporting \(f\) across the homeomorphism), which is why it applies just as well to a coffee cup's cross-section or a crumpled sheet of paper as to a perfect round disc — the popular "stirred coffee" illustration is a genuine instance for \(n=2\), not just an analogy.

A common misconception is that the "no-retraction" fact (Step 6 being impossible) is somehow more elementary or more fundamental than Brouwer itself; in fact the two are formally equivalent (each derivable from the other in a few lines), and neither is provable by purely algebraic or purely point-set-topological means alone — some genuinely algebraic-topological invariant (homology, cohomology, or an equivalent such as degree theory) is unavoidable. This is made precise by the fact that \(D^n\) and \(S^{n-1}\) are indistinguishable by any invariant that does not see "holes", e.g. connectedness, compactness, or metrisability all agree on the two spaces, yet the retraction is still impossible.

Worked examples

Let \(n=2\) and \(f(x,y) = \left(\dfrac{x - y}{2},\ \dfrac{x+y}{2}\right)\) on \(D^2 = \{x^2+y^2\le 1\}\). Show \(f\) has a fixed point without solving for it directly from first principles.

We first check the hypotheses of the theorem apply, then invoke it. A

\(f\) is linear, hence continuous, and \(|f(x,y)|^2 = \tfrac14\big[(x-y)^2+(x+y)^2\big] = \tfrac12(x^2+y^2) \le \tfrac12 \le 1\) whenever \((x,y)\in D^2\), so \(f(D^2)\subseteq D^2\).

Verifying the hypotheses "continuous" and "self-map of \(D^2\)" of the Statement directly by algebra. A

By the theorem, \(\exists (x_0,y_0)\in D^2\) with \(f(x_0,y_0)=(x_0,y_0)\).

Direct application of the Result. A

\[ \frac{x-y}{2}=x,\quad \frac{x+y}{2}=y \;\Longrightarrow\; x=-y,\ y=x \;\Longrightarrow\; x=y=0. \]

Having existence guaranteed, we may now solve the linear system explicitly; a linear map's fixed points form the eigenspace for eigenvalue \(1\), here forced to be \(\{0\}\), confirming the theorem's prediction while also identifying the point exactly (linearity makes this case explicitly solvable, unlike a general continuous \(f\)). A

\((x_0,y_0) = (0,0)\) is the unique fixed point of \(f\) in \(D^2\)

Reading. Brouwer's theorem told us a fixed point had to exist before we did any algebra; the explicit computation then pins it down for this particular linear map.

A continuous demand function \(f:D^2\to D^2\) models price-adjustment on a market with two goods, where \(D^2\) is the (normalised, compact, convex) set of admissible price vectors and \(f(p)\) is the price vector the market moves toward from current prices \(p\). Show an equilibrium price vector exists, given only that \(f\) is continuous and price-preserving in the sense \(f(D^2)\subseteq D^2\).

This is the abstract economic scenario the theorem is classically used to settle; we do not need an explicit formula for \(f\). B

The hypotheses of the Statement are exactly: \(D^2\) is a closed ball (compact, convex — true by normalisation of the price simplex, taken homeomorphic to \(D^2\)), and \(f\) is continuous, self-mapping.

Checking hypotheses against the model's stated properties; note the theorem, via the Corollaries entry on retracts, applies verbatim to anything homeomorphic to \(D^2\), covering the price simplex rather than literally the round disc. B

By the theorem, \(\exists\, p^* \in D^2\) with \(f(p^*) = p^*\).

Direct application of the Result; no further information about \(f\) — not even a formula — was needed. A

An equilibrium price vector \(p^*\) exists

Reading. Purely from continuity and boundedness/convexity of the price space, the market is guaranteed at least one price at which the adjustment process has nothing left to do — this is the essential content of the classical Brouwer-based existence proofs of Walrasian equilibrium.

Problems

Show that Brouwer's theorem for \(n=1\) (every continuous \(f:[-1,1]\to[-1,1]\) has a fixed point) follows from the Intermediate Value Theorem alone, without any homology.
Solution
Let \(g(x)=f(x)-x\), continuous on \([-1,1]\). Since \(f(-1)\in[-1,1]\), \(g(-1)=f(-1)-(-1)=f(-1)+1\ge 0\). Since \(f(1)\in[-1,1]\), \(g(1)=f(1)-1\le 0\). By the Intermediate Value Theorem applied to the continuous function \(g\) on \([-1,1]\), since \(g(-1)\ge0\ge g(1)\), there is \(x_0\in[-1,1]\) with \(g(x_0)=0\), i.e. \(f(x_0)=x_0\). This is the \(n=1\) case of the general theorem, obtainable without the homological machinery of Steps 7–9 because \(H_0(S^0)\cong\mathbb{Z}^2\) and the retraction argument degenerates to exactly this sign-change argument.
Let \(f:D^2\to D^2\) be the map \(f(x,y) = \big(-x,\,-y\big)\) restricted and rescaled by \(\tfrac12\), i.e. \(f(x,y)=(-x/2,-y/2)\). Find all fixed points directly and verify consistency with the theorem.
Solution
Fixed points solve \(-x/2=x\) and \(-y/2=y\), i.e. \(3x/2=0,\ 3y/2=0\), so \((x,y)=(0,0)\) is the only solution. Check hypotheses: \(f\) is linear hence continuous, and \(|f(x,y)| = \tfrac12|(x,y)|\le \tfrac12 \le 1\) on \(D^2\), so \(f(D^2)\subseteq D^2\). The theorem guarantees at least one fixed point; the explicit computation shows there is exactly one, \((0,0)\), consistent with (but stronger than) the theorem's existence-only conclusion.
Explain precisely why the argument of Steps 2–6 breaks down if \(f\) is only required to satisfy \(f(D^n)\subseteq \mathbb{R}^n\) rather than \(f(D^n)\subseteq D^n\), by identifying which step first fails, using \(f(x)=2x\) restricted to \(D^1=[-1,1]\) as the test case away from \(x=0\).
Solution
With \(f(x)=2x\) on \([-1,1]\), \(f(1)=2\notin D^1\), so the hypothesis \(f(D^n)\subseteq D^n\) of the Statement fails outright, and the theorem's conclusion need not hold. Tracing the proof: Step 2 defined \(r(x)\) using \(f(x)\in D^n\) to guarantee, via convexity, that the segment \([f(x),x]\) lies in \(D^n\) (used in Step 5 to pin down \(t=0\) on the boundary). If \(f(x)\) can lie outside \(D^n\), that convexity argument is no longer available, and even though the quadratic in Step 2 might still have a nonnegative root, Step 5 (that \(r\) fixes the boundary) can fail, so \(r\) need not be a genuine retraction and Steps 6–9 no longer apply. Concretely for \(f(x) = 2x\) on \([-1,1]\), the only fixed point candidate solves \(2x=x\), i.e. \(x=0\), which happens to work here — but this is not guaranteed in general once the self-map hypothesis is dropped (the disproof-relevant example in Hypotheses/Fails without uses \(f(x)=x+v,\ v\ne0\), which has no fixed point at all).
Use Brouwer's theorem to show that any continuous function \(g:[0,1]\to[0,1]\) that is also non-decreasing must have a fixed point, and explain whether monotonicity was actually needed.
Solution
\([0,1]\) is homeomorphic to \(D^1=[-1,1]\) via the affine map \(\phi(x)=2x-1\), and \(h = \phi\circ g\circ\phi^{-1}:D^1\to D^1\) is continuous with \(h(D^1)\subseteq D^1\) since \(g([0,1])\subseteq[0,1]\). By the theorem (transported across the homeomorphism, as in the Result's Scope paragraph), \(h\) has a fixed point, and transporting back, \(g\) has a fixed point in \([0,1]\). Monotonicity was not needed at all — continuity and self-mapping of a compact interval already force a fixed point by Problem 1's Intermediate Value Theorem argument; monotonicity is a red herring here (it would matter for uniqueness or for iterative convergence questions, but not for existence).
(Harder.) Give a careful proof, from the Statement, that Brouwer's theorem implies the following version of Kakutani-flavoured self-consistency: if \(A\) is a nonempty compact convex subset of \(\mathbb{R}^n\) (not assumed to be a ball) and \(f:A\to A\) is continuous, then \(f\) has a fixed point. (You may use, without proof, that every nonempty compact convex subset of \(\mathbb{R}^n\) with nonempty interior is homeomorphic to \(D^n\) if bounded with nonempty interior; handle the possibly-lower-dimensional or empty-interior case by a separate short argument.)
Solution
Case 1: \(A\) has nonempty interior (as a subset of \(\mathbb{R}^n\)). By the cited fact, there is a homeomorphism \(\phi:A\to D^n\). Define \(h=\phi\circ f\circ \phi^{-1}:D^n\to D^n\); this is continuous (composite of continuous maps) and maps \(D^n\) into itself since \(f(A)\subseteq A\) and \(\phi\) is a bijection \(A\to D^n\). By the theorem, \(h\) has a fixed point \(z_0\in D^n\), \(h(z_0)=z_0\). Let \(x_0=\phi^{-1}(z_0)\in A\); then \(\phi(f(x_0)) = h(\phi(x_0)) = h(z_0) = z_0 = \phi(x_0)\), and since \(\phi\) is injective, \(f(x_0)=x_0\). Case 2: \(A\) has empty interior in \(\mathbb{R}^n\). Then \(A\) is contained in some affine subspace \(V\cong\mathbb{R}^m\), \(m\lt n\), of \(\mathbb{R}^n\); since \(A\) is compact and convex with nonempty interior relative to \(V\) (a convex set is always full-dimensional in the affine span of itself), apply Case 1 inside \(V\cong\mathbb{R}^m\) instead of \(\mathbb{R}^n\) — the theorem and the cited homeomorphism fact both hold with \(n\) replaced by \(m\), giving the same conclusion. (\(A=\varnothing\) is excluded by hypothesis, so no further case is needed.) This shows the ball-shape hypothesis in the Statement can be replaced by "compact, convex, nonempty" without loss, which is the version of the theorem most often used in applications such as Nash equilibrium existence.