CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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An algebraic complement need not be a topological complement

Statement refuted

Refuted claim: every algebraic direct-sum decomposition of a normed space is automatically a topological direct sum.

Assume the Axiom of Choice and fix a Hamel basis $H$ of $c_{0}$ containing all standard unit vectors $e_{n}$ . Let $M : = span {e_{0}}$ , let $W$ be the kernel of the linear map $P : c_{0} \to M$ defined on basis vectors by

$P (e_{0}) = e_{0}, P (e_{n}) = n e_{0} for n \geq 1, P (h) = 0 for h \in H ∖ {e_{n} : n \geq 0},$

and extend linearly. Then $c_{0} = M \oplus W$ algebraically, but $W$ is not a topological complement of $M$ .

Facts & Assumptions

Given: The normed space $c_{0}$ , the Hamel basis $H$ , the one-dimensional subspace $M = span {e_{0}}$ , and the algebraic projection $P$ above.

[L2]

A topological complement is a direct-sum partner with bounded coordinate projections (A complemented closed subspace of a normed space).

Counterexample

technique · direct

1.1

By construction, $P$ is linear, $P^{2} = P$ , and $ran (P) = M$ . So $W : = \ker P$ satisfies $c_{0} = M \oplus W$ algebraically: every vector decomposes as $P x + (x - P x)$ , and the intersection is trivial because $P$ acts as the identity on $M$ and vanishes on $W$ .

givenalgebra

2.1

The projection $P$ is not bounded for the supremum norm. Indeed, $u_{n} : = e_{n} / n$ satisfies $∥ u_{n} ∥_{\infty} = 1 / n \to 0$ , while $P (u_{n}) = e_{0}$ for every $n \geq 1$ , so $∥ P (u_{n}) ∥_{\infty} = 1$ . No bounded linear map can behave this way at $0$ .

step 1.1

3.1

Suppose, toward a contradiction, that $W$ were a topological complement of $M$ . Then [L2] gives a bounded projection onto $M$ along $W$ , and that projection is unique because the decomposition $x = m + w$ with $m \in M$ , $w \in W$ determines the projection value $m$ pointwise. But $P$ already has exactly that range and kernel by step 1.1, so the bounded projection would have to equal $P$ , contradicting step 2.1.

step 1.1step 2.1L2assume-contradischarge-contradiction

4.1

Therefore $c_{0} = M \oplus W$ is an algebraic decomposition that is not a topological direct sum. This refutes the claim.

step 1.1step 3.1∎

Remarks

The example is intentionally non-load-bearing: it uses a Hamel basis and therefore the Axiom of Choice.
The point is not that complements are rare, but that boundedness of the coordinate projections is extra structure and must be stated.

Depends on

A complemented closed subspace of a normed space

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

4 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Hamel basis (Wikipedia) (standard reference, not scraped)
Haim Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (standard reference, not scraped)