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A closed subspace of ell-infinity that is not complemented

Statement refuted

Assuming the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), the space $c_{0}$ is a closed subspace of $ℓ^{\infty}$ that is not complemented in it: there is no bounded linear projection of $ℓ^{\infty}$ onto $c_{0}$ . Consequently no decomposition $ℓ^{\infty} = c_{0} \oplus E_{1}$ into closed subspaces exists, so the split-chart condition of Split Banach submanifold fails for the pair $(ℓ^{\infty}, c_{0})$ at the identity chart. The ambient $ℓ^{\infty}$ is not second countable and hence is not a Banach manifold in the library's sense (Countable base Banach manifold and smooth map), so this counterexample separates closedness from complementedness at the level of Banach spaces; the sentence about $c_{0}$ in Split Banach submanifold records the same qualification.

Facts & Assumptions

Given: The sequence spaces $c_{0} \subseteq ℓ^{\infty}$ with the sup norm (The sequence spaces c_0 and ell-infinity), the identity chart of $ℓ^{\infty}$ , and the assumed ${A C}_{ω}$ .

[F1]

The sup-normed space $ℓ^{\infty} (K)$ is Banach for $K \in {R, C}$ . Indeed, for a sup-norm Cauchy sequence $(x^{(m)})$ , every coordinate sequence $(x_{n}^{(m)})_{m}$ is Cauchy and has a unique scalar limit $x_{n}$ by real or complex completeness (The reals are complete, The complex plane is complete, and convergence is equivalent to convergence of real and imaginary parts); Replacement collects these unique limits into a sequence $x$ (The Axiom Schema of Replacement: for each formula $φ$ , if $φ$ defines a class function on $A$ then its image on $A$ is a set). Given $ε > 0$ , choose $M$ so that $∥ x^{(m)} - x^{(k)} ∥_{\infty} < ε / 2$ for $m, k \geq M$ ; fixing $m \geq M$ and passing $k \to \infty$ coordinatewise gives $∣ x_{n}^{(m)} - x_{n} ∣ \leq ε / 2$ for every $n$ , so $x$ is bounded and $∥ x^{(m)} - x ∥_{\infty} \leq ε / 2$ . Thus every Cauchy sequence converges in $ℓ^{\infty}$ , as required by Banach space. The subspace $c_{0}$ is closed in $ℓ^{\infty}$ ( $c_{0}$ is a closed subspace of $ℓ^{\infty}$ ) and therefore Banach by the closed-subspace theorem (A closed subspace of a Banach space is Banach); this also agrees with the direct result Real and complex $c_{0}$ are Banach.

[L1]

The identity chart of $ℓ^{\infty}$ covers $ℓ^{\infty}$ and has trivial transition maps, so the split-chart condition of Split Banach submanifold is meaningful for the pair $(ℓ^{\infty}, c_{0})$ : it asks for a decomposition $ℓ^{\infty} = E_{0} \oplus E_{1}$ into closed subspaces with bounded coordinate projections such that, in the chart, $U \cap c_{0} = U \cap (E_{0} \oplus {0})$ . The space $ℓ^{\infty}$ is not second countable — the uncountably many $0$ - $1$ sequences are pairwise at sup-distance $1$ , so every dense subset is uncountable — hence $ℓ^{\infty}$ is not a Banach manifold in the library's sense (Countable base Banach manifold and smooth map, Second countability: an at most countable basis for the topology).

[L2]

A bounded linear projection $P$ of $ℓ^{\infty}$ onto $c_{0}$ fixes every element of $c_{0}$ , and for each $n$ the map $x \mapsto x_{n} - (P x)_{n}$ is a bounded linear functional on $ℓ^{\infty}$ that annihilates $c_{0}$ , hence induces a bounded linear functional on the quotient $Q = ℓ^{\infty} / c_{0}$ of norm at most $1 + ∥ P ∥$ (A bounded linear operator between normed spaces, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A complemented closed subspace of a normed space, The quotient vector space $X / M$ , its cosets, and the quotient map $q : X \to X / M$ , The quotient seminorm $∥ x + M ∥_{X / M} = \inf_{m \in M} ∥ x + m ∥ = dist (x, M)$ , The dual space X^* of a normed space and its dual norm).

[L3]

The rationals are dense in $R$ and the irrationals are uncountable (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable, The irrationals are uncountable).

[L4]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

[L5]

Under ${A C}_{ω}$ , a countable union of countable sets is countable (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Countable unions of at most countable sets, assuming ${A C}_{ω}$ ).

[L6]

Cosets of the quotient $Q = ℓ^{\infty} / c_{0}$ , the quotient map, the quotient seminorm $∥ x + c_{0} ∥_{Q} = dist (x, c_{0})$ , and its being a norm because $c_{0}$ is closed (The quotient vector space $X / M$ , its cosets, and the quotient map $q : X \to X / M$ , The quotient seminorm $∥ x + M ∥_{X / M} = \inf_{m \in M} ∥ x + m ∥ = dist (x, M)$ , The quotient seminorm is a norm exactly when the subspace is closed).

[L7]

A closed subspace is complemented exactly when it is the range of a bounded linear projection (A closed subspace is complemented exactly when it is the range of a bounded projection, A complemented closed subspace of a normed space).

[L8]

Dual and operator bounds: $∣ g (v) ∣ \leq ∥ g ∥ ∥ v ∥$ for $g$ in the dual, and $∥ P x ∥ \leq ∥ P ∥ ∥ x ∥$ for a bounded linear $P$ (The dual space X^* of a normed space and its dual norm, The operator norm as the least bound and as the unit-sphere or unit-ball supremum, A bounded linear operator between normed spaces).

Counterexample

technique · contradiction

1.1F1L1

$c_{0}$ is closed in $ℓ^{\infty}$ and is a Banach space for the sup norm by [F1]; $ℓ^{\infty}$ is not second countable and hence is not a Banach manifold in the library's sense, while the identity chart still makes the split-chart condition meaningful for the pair $(ℓ^{\infty}, c_{0})$ by [L1].

1.2assume-contraL7

Suppose for contradiction that $c_{0}$ is complemented in $ℓ^{\infty}$ : by [L7] fix a bounded linear projection $P : ℓ^{\infty} \to c_{0}$ of $ℓ^{\infty}$ onto $c_{0}$ , so that $P x = x$ for every $x \in c_{0}$ .

1.3L3L4construct

Fix an enumeration $(q_{n})$ of the rationals; for every irrational $x$ and every $k \geq 1$ let $n_{k} (x)$ be the least index $n$ not already used among $n_{1} (x), \dots, n_{k - 1} (x)$ with $q_{n} \in (x - \frac{1}{k}, x + \frac{1}{k})$ , and put $A_{x} : = {n_{k} (x) : k \geq 1}$ .

2.1step 1.3L3algebra

Each $A_{x}$ is infinite. If $x \neq y$ , choose $K$ so large that the intervals $(x - \frac{1}{k}, x + \frac{1}{k})$ and $(y - \frac{1}{ℓ}, y + \frac{1}{ℓ})$ are disjoint whenever $k, ℓ \geq K$ . Any common index of $A_{x}$ and $A_{y}$ must therefore occur among the first $K - 1$ choices for at least one of $x, y$ , a finite set; hence $A_{x} \cap A_{y}$ is finite. It follows that $x \mapsto A_{x}$ is injective and ${A_{x}}$ is uncountable by [L3].

3.1step 1.3step 2.1F1L6algebra

Let $u_{x}$ be the indicator sequence of $A_{x}$ and let $Q : = ℓ^{\infty} / c_{0}$ carry the quotient norm; then $u_{x} \in ℓ^{\infty}$ and $[u_{x}] \neq 0$ because $u_{x} \notin c_{0}$ . For distinct $x_{1}, \dots, x_{m}$ , remove the finite union of all pairwise intersections from their supports. The resulting indicators have disjoint supports and differ from the $u_{x_{j}}$ by finitely supported, hence $c_{0}$ , sequences. Therefore, for scalars $c_{1}, \dots, c_{m}$ , the quotient norm of $\sum_{j} c_{j} [u_{x_{j}}]$ equals $\max_{j} ∣ c_{j} ∣$ : the disjoint representative gives the upper bound, and each remaining infinite support attains the corresponding coefficient infinitely often, giving the lower bound against every $c_{0}$ perturbation.

4.1step 3.1L8choosealgebra

For every $g \in Q^{*}$ and every real $r > 0$ the set ${x : ∣ g ([u_{x}]) ∣ \geq r}$ is finite: for distinct points $x_{1}, \dots, x_{m}$ in it choose unimodular scalars $c_{j}$ with $c_{j} g ([u_{x_{j}}]) = ∣ g ([u_{x_{j}}]) ∣$ , so that by [step 3.1] and [L8] one has $m r \leq \sum_{j} ∣ g ([u_{x_{j}}]) ∣ = ∣ g (\sum_{j} c_{j} [u_{x_{j}}]) ∣ \leq ∥ g ∥ \cdot ∥ \sum_{j} c_{j} [u_{x_{j}}] ∥_{Q} = ∥ g ∥$ and hence $m \leq ∥ g ∥ / r$ .

5.1step 4.1step 2.1L5

No countable family in $Q^{*}$ separates the points of $Q$ : given $(g_{n})$ , the set of $x$ with $g_{n} ([u_{x}]) \neq 0$ for some $n$ is the countable union over the pairs $(n, k)$ of the finite sets of [step 4.1] with $r = \frac{1}{k}$ , hence countable by [L5]; since ${A_{x}}$ is uncountable by [step 2.1], some $x$ lies outside it, and then the nonzero vector $[u_{x}]$ is annihilated by every $g_{n}$ .

6.1step 1.2step 5.1L2L6L8

Let $P$ be the projection assumed in step 1.2 and put $ψ_{n} (x + c_{0}) : = x_{n} - (P x)_{n}$ . By [L2] each $ψ_{n}$ is a well-defined bounded linear functional on $Q$ — well-defined because $P$ fixes every element of $c_{0}$ — with $∥ ψ_{n} ∥ \leq 1 + ∥ P ∥$ by [L8], and if $ψ_{n} (x + c_{0}) = 0$ for all $n$ then $(x - P x)_{n} = 0$ for all $n$ , so $x = P x \in c_{0}$ . The countably many functionals $ψ_{n}$ would therefore be a countable separating family in $Q^{*}$ , contradicting [step 5.1].

7.1step 1.1step 5.1step 6.1L7discharge-contradiction∎

This contradiction with [step 5.1] shows that no bounded linear projection of $ℓ^{\infty}$ onto $c_{0}$ exists; by [L7] $c_{0}$ is not complemented in $ℓ^{\infty}$ , although it is closed there by [step 1.1], and consequently the identity chart admits no split-chart decomposition of $c_{0}$ , as claimed.

Remarks

Where the countability enters. The proof only uses ${A C}_{ω}$ once, in [step 5.1], to make the union of the finitely-many-violators sets countable. The construction of the uncountable family ${A_{x}}$ and the quotient-norm computation are choice-free beyond the fixed enumeration of the rationals.
The manifold reading. The identity chart makes $(ℓ^{\infty}, c_{0})$ an instance of the split-chart condition, but $ℓ^{\infty}$ is not second countable, so the pair is not a Banach manifold. The library's Split Banach submanifold therefore treats this example as evidence that closedness does not imply splitness in the Banach-space setting, and the split-submanifold definition itself is stated only for second countable ambient manifolds.

Depends on

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Sources

Piotr Hajłasz, Functional Analysis — Theorem 10.19 (c_0 is not complemented in ℓ∞) (standard reference, not scraped)