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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01
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The quotient seminorm is a norm exactly when the subspace is closed

Statement

Let X be a normed space and let M≤X. The quotient seminorm on X/M is a norm if and only if M is closed in X.

Facts & Assumptions

Given: A normed space X, a linear subspace M≤X, and a vector x∈X.

[L1]

The quotient seminorm is ∥x+M∥X/M=dist⁡(x,M) (The quotient seminorm (|x+M|{X/M}=\inf{m\in M}|x+m|=\operatorname{dist}(x,M))).

[L2]

A linear subspace contains 0, so M is nonempty (Linear subspace of a vector space).

[L3]

For a nonempty subset A of a metric space, the closure of A is exactly {u:dist⁡(u,A)=0}, and a set is closed exactly when it equals its closure (The closure of a nonempty A is {x:d(x,A)=0}, equals A together with its limit points, and is the smallest closed superset).

Proof

technique · direct
1.1L1L2L3

By [L2] and [L3], dist⁡(x,M)=0 exactly when x∈M‾. Therefore [L1] gives ∥x+M∥X/M=0 exactly when x∈M‾.

2.1step 1.1L3

If M is closed and ∥x+M∥X/M=0, then step 1.1 gives x∈M‾=M. Hence x+M=M, the zero coset. So the quotient seminorm is definite and therefore a norm.

2.2step 1.1L3assume-contradischarge-contradiction

Conversely, assume the quotient seminorm is a norm. If M were not closed, then [L3] would give some x∈M‾∖M. Step 1.1 would then give ∥x+M∥X/M=0, while x+M≠M because x∉M, contradicting definiteness. Therefore M is closed.

3.1step 2.1step 2.2∎

Steps 2.1 and 2.2 prove the equivalence.

Depends on

Used by

Dependency tree · two levels

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Sources