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

1.1

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

L1L2L3choose

2.1

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

step 1.1L1

3.1

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

step 2.1L3∎

Depends on

Used by

A bounded operator that vanishes on a subspace factors uniquely through the normed quotient Theorem

Dependency tree · two levels

10 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

Theo Buhler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)
Gerald Teschl, Topics in Real and Functional Analysis (standard reference, not scraped)