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

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The quotient map sends every open ball onto a set containing the corresponding quotient ball

Statement

Let $X$ be a normed space, let $M \leq X$ be closed, and let $q : X \to X / M$ be the quotient map. Then for every $x_{0} \in X$ and every $r > 0$ ,

$q (B_{X} (x_{0}, r)) \supseteq B_{X / M} (q (x_{0}), r) .$

In particular, $q$ is an open map.

Facts & Assumptions

Given: A normed space $X$ , a closed linear subspace $M \leq X$ , a vector $x_{0} \in X$ , a real $r > 0$ , and a coset $ξ \in X / M$ .

[L1]

The quotient map is $q (x) = x + M$ , and addition of cosets is inherited from the vector-space quotient (The quotient vector space (X/M), its cosets, and the quotient map (q:X\to X/M)).

[L2]

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

[L3]

Because $M$ is closed, the quotient seminorm is an honest norm on $X / M$ (The quotient seminorm is a norm exactly when the subspace is closed).

Proof

technique · direct

1.1L1L2L3choose

First take $x_{0} = 0$ . Let $ξ = x + M$ satisfy $∥ ξ ∥_{X / M} < r$ . By [L2], choose $m \in M$ with $∥ x + m ∥ < r$ . Then $x + m \in B_{X} (0, r)$ and $q (x + m) = x + m + M = x + M = ξ$ by [L1]. Hence $B_{X / M} (0, r) \subseteq q (B_{X} (0, r))$ .

2.1step 1.1L1

For general $x_{0}$ , a coset $ξ$ lies in $B_{X / M} (q (x_{0}), r)$ exactly when $ξ - q (x_{0}) \in B_{X / M} (0, r)$ . By step 1.1 there is $u \in B_{X} (0, r)$ with $q (u) = ξ - q (x_{0})$ . Then $q (x_{0} + u) = q (x_{0}) + q (u) = ξ$ , and $x_{0} + u \in B_{X} (x_{0}, r)$ . Therefore $q (B_{X} (x_{0}, r)) \supseteq B_{X / M} (q (x_{0}), r)$ .

3.1step 2.1L3∎

Every open ball in $X$ has image containing an open ball in $X / M$ , so $q$ is open.

The quotient map sends every open ball onto a set containing the corresponding quotient ball

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources