Separable Quotient Problem for Banach Spaces

Separable Quotient Problem for Banach Spaces

OpenAI's claim

OpenAI presents the following claims relevant to this question, with manuscript numbers following paper order in Sources. In manuscript 1 (family 323; 2026-09-23), the asserted result is that the general separable-quotient assertion is independent of ZFC relative to a measurable cardinal, with a ZFC-only relative-independence result at density $\aleph_1$.

Mathematical statement of the claimed result

Manuscript 1.

For $\mathbb F=\mathbb R$ or $\mathbb C$, write $\mathrm{SQ}_{\mathbb F}$ for the following assertion: each Banach space of infinite dimension over $\mathbb F$ admits a bounded linear quotient whose range is separable and remains infinite dimensional. In ZFC, the manuscript claims \[ \mathfrak c\text{ real-valued measurable}\Rightarrow\mathrm{SQ}_{\mathbb F}, \qquad\mathrm{CH}\Rightarrow\neg\mathrm{SQ}_{\mathbb F}. \] Here real-valued measurable means a probability measure on every subset of the cardinal $\mathfrak c=2^{\aleph_0}$, vanishing on singletons and additive on disjoint families of size less than $\mathfrak c$. If ZFC plus a measurable cardinal is consistent, the separable quotient assertion is therefore claimed independent of ZFC. Relative consistency of its failure needs only consistency of ZFC.

Outline of the argument

For manuscript 1 (family 323; 2026-09-23), a coded coordinate predual uses norming paths to prevent separable evaluation images in the counterexample model. Block restrictions and strictly singular estimates establish the obstruction, while a continuum measure hypothesis supplies the positive-model direction.

Formal verification

For manuscript 1 (family 323; 2026-09-23), the documented Lean scope is under CH, real and complex Banach counterexamples with no infinite-dimensional separable quotient; the positive-model direction and complete relative-independence assertion are outside the selected theorem.

This coverage does not establish a formal proof of the complete problem as stated here.

The linked family Comparator checks are SeparableQuotientNegative.lean.

Their named declarations include OAI.SeparableQuotient.negative_main.

This site has not executed the Lean code. Described statement coverage is not an independent proof check.

Status caveat

The manuscripts were produced by an internal OpenAI model. OpenAI cautions that results without formalization can contain mistakes. This site has not independently checked the proofs; the label records the scope of OpenAI's claim.

Sources

• OpenAI, Relative independence of the separable quotient problem, 2026-09-23; citation key OAI:Relative-independence-of-the-separable-quotient-problem-September-23-2026. PDF (GitHub); PDF (direct); Manuscript page and citation; Latest version; Statement source: intro:main.
• Family 323: Lean scope.
• Comparator SeparableQuotientNegative.json.
• Comparator SeparableQuotientNegative.lean.
• OpenAI catalogue entry.

Separable Quotient Problem for Banach Spaces

Definitions and mathematical statement

A Banach space is a complete normed real or complex vector space. For a closed linear subspace $Y\subset X$, the quotient $X/Y$ carries the complete quotient norm \[ \|x+Y\|_{X/Y}=\inf_{y\in Y}\|x-y\|_X. \] A space is separable if it has a countable dense subset. The separable quotient problem asks whether every infinite-dimensional Banach space $X$ has some closed $Y$ for which $X/Y$ is both infinite-dimensional and separable. Neither a finite-dimensional quotient nor a separable subspace of $X$ answers this quotient question. For an already separable infinite-dimensional $X$, the zero subspace is allowed; the universal statement includes nonseparable spaces without a chosen basis or additional geometric structure.

Formulation and concept references: [S1].

Short English statement

Does every infinite-dimensional Banach space admit an infinite-dimensional separable quotient by a closed subspace?

Research attempt

Scope, attribution, and source review.

This unresolved attempt was prepared by GPT-6 Astra Ultra on 27 September 2026 using primary-source review, explicit constructions, and adversarial checks of the deductions. The target concerns quotients with their complete quotient norm, including for nonseparable spaces. Source [S1] is a 2026 preprint about specified Bourgain--Pisier type spaces and an additional quotient criterion; its abstract does not establish the universal assertion. No result claimed there is needed below. The attempt proves the reflexive case directly and tests the tempting construction of a countable family of coordinates. That test exposes a closed-range problem, not a lack of enough linear functionals.

A complete subcase: reflexive spaces.

Let $X$ be an infinite-dimensional reflexive Banach space, over either $\mathbb R$ or $\mathbb C$. Choose linearly independent functionals $f_1,f_2,\ldots\in X^*$ and put \[ Z=\overline{\operatorname{span}}\{f_j:j\geq1\}\subset X^*. \] The space $Z$ is separable, infinite-dimensional and reflexive. Define $Q:X\to Z^*$ by $Qx(z)=z(x)$. It is bounded with norm at most one. It is also onto: a functional $\phi\in Z^*$ extends by Hahn--Banach to $\widetilde\phi\in X^{**}$ with the same norm, and reflexivity writes $\widetilde\phi(z)=z(x)$ for some $x\in X$. Thus $Qx=\phi$.

Here $Z^*$ is separable. To check this sometimes omitted point, the unit ball of $Z^*$ is weak-star compact and metrizable because $Z$ is separable. Reflexivity makes its weak and weak-star topologies agree. Take a countable weakly dense subset of that ball. Its rational convex combinations (using real rational coefficients even in the complex case) are countable and norm dense in the ball: otherwise Hahn--Banach separation would separate a point from their closed convex hull, contradicting weak density. Consequently $Z^*$ is norm separable.

The kernel \[ Y=Z^\perp=\{x:z(x)=0\text{ for every }z\in Z\} \] is closed. In fact the induced map $X/Y\to Z^*$ is an isometry, since the norm-preserving extension above shows that every $\phi$ has a representative $x$ with $\|x\|=\|\phi\|$, while the reverse inequality follows from $\|Q\|\leq1$. This gives the required infinite-dimensional separable quotient for every reflexive $X$. It includes nonseparable Hilbert spaces. Reflexivity is used exactly to turn the Hahn--Banach extension in $X^{**}$ into an element of $X$; deleting that step does not prove the general problem.

Other elementary families.

If $X$ has a complemented infinite-dimensional separable subspace $E$, a bounded projection $P:X\to E$ gives $X/\ker P\cong E$. An arbitrary separable subspace need not be complemented, so merely taking the closed span of a countable independent sequence does not give this argument. For $\ell^p(\Gamma)$, $1\leq p<\infty$, and for $c_0(\Gamma)$, restriction to a countably infinite subset $\Gamma_0\subset\Gamma$ is onto the corresponding separable sequence space. Extension by zero is a bounded right inverse. These assertions verify the quotient norm and surjectivity, rather than just a coordinatewise map into a separable ambient space.

Countably many coordinates always give a dense-range map.

Every infinite-dimensional Banach space admits a biorthogonal sequence $(x_j,f_j)$ with $f_j(x_i)=\delta_{ij}$. Here is an induction that needs no basis for $X$. Having chosen finitely many pairs, their common functional kernel is infinite-dimensional. Choose a nonzero vector there outside the finite span of previous vectors, and use Hahn--Banach on that finite-dimensional span plus the new vector to produce the next functional vanishing on previous vectors. Normalize the new pair to have value one. This maintains biorthogonality in both directions.

Choose $a_j>0$ with $\sum_j a_j^2\|f_j\|^2<\infty$, for example $a_j=2^{-j}/(1+\|f_j\|)$. Then \[ Tx=(a_jf_j(x))_{j\geq1} \] defines a bounded operator $T:X\to\ell^2$. The image contains every finitely supported sequence, because $Tx_j=a_je_j$. Its range is therefore dense and infinite-dimensional. Nevertheless $T$ is compact: truncating at coordinate $N$ gives a finite-rank operator and \[ \|T-T_N\|\leq \left(\sum_{j>N}a_j^2\|f_j\|^2\right)^{1/2}\longrightarrow0. \] It cannot be onto $\ell^2$. If it were onto, the open mapping theorem would put a fixed ball of $\ell^2$ inside the image of a bounded ball of $X$, contradicting relative compactness and the noncompactness of the infinite-dimensional unit ball.

This construction does yield an infinite-dimensional quotient $X/\ker T$, but it does not show that this quotient is separable in its quotient norm. The induced bijection onto $T(X)$ need not have continuous inverse when $T(X)$ is given its inherited $\ell^2$ norm. That distinction is exactly where the naive argument fails.

An explicit norm-topology countertest.

Take $X=\ell^\infty$ and \[ Tx=(2^{-j}x_j)_{j\geq1}\in\ell^2. \] The map is bounded, compact and injective, and its range contains all finite sequences. Thus the kernel is zero and the quotient is $\ell^\infty$ itself, which is nonseparable. One elementary proof is that all indicator sequences of subsets of $\mathbb N$ form an uncountable set of pairwise distance one. The separable completion of $T(X)$ in the weaker target norm has not become a separable Banach quotient of the original space.

There is also a concrete failure of surjectivity: the sequence $y_j=2^{-j}\sqrt j$ lies in $\ell^2$, while its only possible preimage has coordinates $\sqrt j$ and is unbounded. This example does not say that $\ell^\infty$ has no separable quotient. It says that this particular countable-coordinate construction has not produced one.

A quantitative criterion that would repair the construction.

Suppose $(g_j)\subset X^*$ is bounded and weak-star null, so $g_j(x)\to0$ for every $x$. Then \[ Qx=(g_j(x))_{j\geq1} \] is a bounded map into $c_0$. Its adjoint on $\ell^1=c_0^*$ is $Q^*a=\sum_j a_jg_j$, with norm convergence. A sufficient condition for surjectivity is \begin{equation}\label{wsref-0-61-34-30-35-3a-6c-6f-77-65-72} \left\|\sum_j a_jg_j\right\|\geq c\sum_j|a_j| \quad\text{for all finitely supported }a,\qquad c>0. \end{equation} The inequality extends to all $\ell^1$. To see why it gives onto, Hahn--Banach separation shows that the norm closure of $Q(B_X)$ contains $cB_{c_0}$: its support function at $a\in\ell^1$ is $\|Q^*a\|\geq c\|a\|_1$. For complex spaces apply the same real separation argument with real parts of complex functionals.

Given $y\in c_0$, approximate it by $Qx_1$ with $\|x_1\|\leq2\|y\|/c$ and error at most $\|y\|/2$. Repeat for the residual with successive errors halved. The resulting series $\sum x_j$ converges in $X$, and its image is exactly $y$. This proves surjectivity and hence the desired quotient when (a405:lower) is available.

Weak-star nullity and normalization alone do not imply that lower bound. In $X=\ell^2$, coordinate functionals have norm one and are weak-star null, but the norm of a sum of the first $N$ is $\sqrt N$, rather than a fixed multiple of $N$. The resulting map $\ell^2\to c_0$ is not onto. This does not contradict the reflexive-case proof, whose quotient need not be $c_0$.

Verification, gap, and outcome.

The proofs identify the topology of each image, the closedness of each kernel and the exact use of completeness. The compact diagonal map and the coordinate-functional test check two different failure modes. No numerical test can replace the universal lower estimate (a405:lower) or the reflexivity argument. The first remaining gap is a mechanism giving an appropriate closed-range map into some infinite-dimensional separable Banach space for an arbitrary nonreflexive $X$. Requiring the particular target $c_0$ would be stronger and is not imposed by the problem. UNRESOLVED. The reflexive and complemented-subspace cases are proved, and a proposed universal shortcut is disproved. No general solution or novelty is claimed.

Sources

• S1 Mazur's Separable Quotient Problem for Nonseparable Bourgain-Pisier spaces. https://arxiv.org/abs/2604.11832. Formulation source.