Topology of infinite-dimensional Banach and Hilbert spaces

DefinitionScottish Book Problem 167#✓fully proved

Ulam's claim that for every infinite-dimensional real Hilbert space \(H\) with unit sphere \(S = \{p \in H : \|p\| = 1\}\), every \(n \in \mathbb{N}\), all continuous functions \(f_1, \dots, f_n : S \to \mathbb{R}\) and every continuous map \(T : S \to S\), there is a point \(p_0 \in S\) with \(f_i(T(p_0)) = f_i(p_0)\) for every \(i\).

def ScottishBookProblem167Claim : Prop :=
  ∀ (E : Type) [NormedAddCommGroup E] [InnerProductSpace ℝ E] [CompleteSpace E],
    ¬ Module.Finite ℝ E →
    ∀ (n : ℕ) (f : Fin n → ↥(Metric.sphere (0 : E) 1) → ℝ)
      (T : ↥(Metric.sphere (0 : E) 1) → ↥(Metric.sphere (0 : E) 1)),
      Continuous T → (∀ i, Continuous (f i)) →
      ∃ p0 : ↥(Metric.sphere (0 : E) 1), ∀ i, f i (T p0) = f i p0

Used by 1

Used byNegative answer to Scottish Book Problem 167

TheoremNegative answer to Scottish Book Problem 167#●ready to prove

Ulam's claim in Problem 167 is false: there are an infinite-dimensional real Hilbert space \(H\) with unit sphere \(S\), a number \(n \in \mathbb{N}\), continuous functions \(f_1, \dots, f_n : S \to \mathbb{R}\) and a continuous map \(T : S \to S\) such that no point \(p_0 \in S\) satisfies \(f_i(T(p_0)) = f_i(p_0)\) for every \(i\).

The claim fails already for one function.

theorem_wanted ulam167_refuted : ¬ ScottishBookProblem167Claim

Sources

• S. Ulam, Problem 167 of the Scottish Book; refuted via V. Klee's homeomorphism of the Hilbert sphere onto the whole Hilbert space: composing that homeomorphism with a translation gives a continuous T and a continuous f with no point p0 satisfying f (T p0) = f p0. See the commentary in R. D. Mauldin (ed.), The Scottish Book, 2nd ed. (Birkhäuser, 2015).

Uses 1

Statement usesScottish Book Problem 167

DefinitionProjective Hilbert space#✓fully proved

Let \(E\) be a normed additive commutative group, in practice a real inner product space. The projective Hilbert space of \(E\) is the quotient of the unit sphere \(\{x \in E : \|x\| = 1\}\) by the relation identifying each \(x\) with \(-x\). Its points are the antipodal pairs \(\{x, -x\}\) of unit vectors, that is, the diameters of the unit sphere.

When \(E\) is a real inner product space, it carries the quotient topology of the unit sphere.

def ProjectiveHilbertSpace (E : Type) [NormedAddCommGroup E] : Type :=
  Quot (fun x y : ↥(Metric.sphere (0 : E) 1) => (x : E) = y ∨ (x : E) = -(y : E))

Used by 1

Used byProjective Hilbert space homeomorphic to the space itself

DefinitionProjective Hilbert space homeomorphic to the space itself#✓fully proved

For a real inner product space \(E\), the statement that the projective Hilbert space of \(E\), the unit sphere \(\{x \in E : \|x\| = 1\}\) with each \(x\) identified with \(-x\) and given the quotient topology, is homeomorphic to \(E\).

def ProjectiveHilbertSpaceHomeomorphicToHilbertSpace (E : Type)
    [NormedAddCommGroup E] [InnerProductSpace ℝ E] : Prop :=
  Nonempty (ProjectiveHilbertSpace E ≃ₜ E)

Uses 1 · Used by 1

Statement usesProjective Hilbert space

Used byProjective Hilbert space is not homeomorphic to Hilbert space

TheoremProjective Hilbert space is not homeomorphic to Hilbert space#●ready to prove

Let \(E\) be an infinite-dimensional real Hilbert space. Then the projective Hilbert space of \(E\), the unit sphere \(\{x \in E : \|x\| = 1\}\) with each \(x\) identified with \(-x\) and given the quotient topology, is not homeomorphic to \(E\).

This answers Ulam's Problem 35 of the Scottish Book in the negative. The projective space is the standard model of the space of diameters of the unit sphere with the Hausdorff metric. Hilbert space is contractible, hence simply connected, while projective Hilbert space is not, its fundamental group being \(\mathbb{Z}/2\) via the double covering by the unit sphere.

theorem_wanted projective_hilbert_space_not_homeomorphic_to_hilbert_space :
    ∀ (E : Type) [NormedAddCommGroup E] [InnerProductSpace ℝ E] [CompleteSpace E],
      ¬ Module.Finite ℝ E →
      ¬ ProjectiveHilbertSpaceHomeomorphicToHilbertSpace E

Sources

• S. Ulam, Problem 35 of the Scottish Book; see the commentary of Holsztynski in R. D. Mauldin (ed.), The Scottish Book, 2nd ed. (Birkhäuser, 2015), and the Hilbert-manifold classification by homotopy type.

Uses 1

Statement usesProjective Hilbert space homeomorphic to the space itself

TheoremRetraction onto the unit sphere of Hilbert space#●ready to prove

Let \(E\) be an infinite-dimensional real Hilbert space and let \(S = \{x \in E : \|x\| = 1\}\) be its unit sphere. Then there is a continuous map \(f : E \to E\) such that \(f(x) \in S\) for every \(x \in E\) and \(f(x) = x\) for every \(x \in S\).

Restricting \(f\) to the closed unit ball gives a retraction of the ball onto \(S\); conversely, a retraction \(r\) of the closed unit ball onto \(S\) extends to all of \(E\) by \(f(x) = r(x / \max(1, \|x\|))\). In finite dimensions no such map exists, by the Brouwer no-retraction theorem, so the infinite-dimensional hypothesis is essential.

theorem_wanted hilbertBallRetractsOntoSphere :
    ∀ (E : Type u) [NormedAddCommGroup E] [InnerProductSpace ℝ E] [CompleteSpace E],
      ¬ Module.Finite ℝ E →
      ∃ f : E → E, Continuous f ∧ (∀ x, f x ∈ Metric.sphere (0 : E) 1) ∧
        ∀ x ∈ Metric.sphere (0 : E) 1, f x = x

Sources

• S. Ulam, Problem 36 of the Scottish Book, asking whether there is a retraction of the closed unit ball of Hilbert space onto its boundary; the answer is affirmative, see the commentary in R. D. Mauldin (ed.), The Scottish Book, 2nd ed. (Birkhäuser, 2015), and the 2026 survey "90+ years of the Scottish Book" (arXiv:2603.27867), which notes that the unit sphere of a Hilbert space is homeomorphic to the whole space.

DefinitionScottish Book Problem 7 (Mazur–Banach)#✓fully proved

The claim that in every real Banach space \(E\), any two closed convex subsets \(A, B \subseteq E\) whose linear spans are infinite-dimensional are homeomorphic, with the subspace topologies.

def ScottishBookProblem7Claim : Prop :=
  ∀ (E : Type) [NormedAddCommGroup E] [NormedSpace ℝ E] [CompleteSpace E]
    (A B : Set E), IsClosed A → Convex ℝ A →
    ¬ Module.Finite ℝ ↥(Submodule.span ℝ A) → IsClosed B → Convex ℝ B →
    ¬ Module.Finite ℝ ↥(Submodule.span ℝ B) → Nonempty (↥A ≃ₜ ↥B)

Used by 1

Used byScottish Book Problem 7 fails: closed convex sets need not be homeomorphic

TheoremScottish Book Problem 7 fails: closed convex sets need not be homeomorphic#●ready to prove

The Scottish Book Problem 7 claim is false: there are a real Banach space \(E\) and closed convex subsets \(A, B \subseteq E\), each with infinite-dimensional linear span, that are not homeomorphic.

For example, a compact infinite-dimensional closed convex set, such as the Hilbert cube, is not homeomorphic to a noncompact one, such as the whole space, since compactness is a topological invariant.

theorem_wanted closedConvexSetsHomeomorphic_refuted : ¬ ScottishBookProblem7Claim

Sources

• S. Mazur and S. Banach, Problem 7 of the Scottish Book; T. Banakh's observation as reported in the 2026 survey "90+ years of the Scottish Book" (arXiv:2603.27867). The modified classification problem was solved for Frechet spaces by Banakh-Cauty, Studia Mathematica 205 (2011) 1-11.

Uses 1

Statement usesScottish Book Problem 7 (Mazur–Banach)

4 fully proved4 ready to prove

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