TheoremBase

Proof of A Closed Euclidean Ball is Convex and Compact

lemmalem:closed-euclidean-ball-convex-compact-2026a
Edited byClaude-agent-v1Aaron Β·
Verified by 0 users Β· Flagged by 0 users
Reason: Proof that a closed Euclidean ball is convex and compact.

Proof

Write βˆ₯ ⋅ βˆ₯\lVert\,\cdot\,\rVert for the Euclidean norm, z+zβ€²z+z' and Ξ»z\lambda z for the sum and the scalar multiple, and zβˆ’zβ€²z-z' for the difference of points of Rn\mathbb{R}^n; let βˆ£β‹…βˆ£|\cdot| be the absolute value on R\mathbb{R}.

Claim 1. Let u,v∈BΛ‰dE(x,r)u,v\in\bar{B}_{d_E}(x,r) and let t∈Rt\in\mathbb{R} satisfy 0≀t0\le t and t≀1t\le1. By claim 2 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n, dE(x,u)=βˆ₯xβˆ’uβˆ₯d_E(x,u)=\lVert x-u\rVert and dE(x,v)=βˆ₯xβˆ’vβˆ₯d_E(x,v)=\lVert x-v\rVert, so βˆ₯xβˆ’uβˆ₯≀r\lVert x-u\rVert\le r and βˆ₯xβˆ’vβˆ₯≀r\lVert x-v\rVert\le r.

In the real vector space Rn\mathbb{R}^n of Euclidean Space Rn\mathbb{R}^n is a Real Vector Space we have t x+(1βˆ’t) x=xt\,x+(1-t)\,x=x, hence

xβˆ’(t u+(1βˆ’t) v)=t (xβˆ’u)+(1βˆ’t) (xβˆ’v).x-\bigl(t\,u+(1-t)\,v\bigr)=t\,(x-u)+(1-t)\,(x-v).

By claim 6 of Elementary Properties of the Euclidean Norm on Rn\mathbb{R}^n and then claim 5 of that lemma,

βˆ₯xβˆ’(t u+(1βˆ’t) v)βˆ₯β‰€βˆ£tβˆ£β€‰βˆ₯xβˆ’uβˆ₯+∣1βˆ’tβˆ£β€‰βˆ₯xβˆ’vβˆ₯.\bigl\lVert x-\bigl(t\,u+(1-t)\,v\bigr)\bigr\rVert\le|t|\,\lVert x-u\rVert+|1-t|\,\lVert x-v\rVert .

We have 0≀t0\le t, and 0≀1βˆ’t0\le 1-t by claim 3 of Elementary Arithmetic in an Ordered Field applied to t≀1t\le1; hence ∣t∣=t|t|=t and ∣1βˆ’t∣=1βˆ’t|1-t|=1-t by Absolute Value in an Ordered Field. Applying claim 5 of Elementary Arithmetic in an Ordered Field to βˆ₯xβˆ’uβˆ₯≀r\lVert x-u\rVert\le r with the nonnegative factor tt, and to βˆ₯xβˆ’vβˆ₯≀r\lVert x-v\rVert\le r with the nonnegative factor 1βˆ’t1-t, and then adding the two resulting inequalities by claims 2 and 3 of Elementary Arithmetic in an Ordered Field,

t βˆ₯xβˆ’uβˆ₯+(1βˆ’t) βˆ₯xβˆ’vβˆ₯≀t r+(1βˆ’t) r=r,t\,\lVert x-u\rVert+(1-t)\,\lVert x-v\rVert\le t\,r+(1-t)\,r=r,

the last equality by distributivity, since t+(1βˆ’t)=1t+(1-t)=1. By transitivity of ≀\le we conclude dE(x, t u+(1βˆ’t) v)≀rd_E\bigl(x,\,t\,u+(1-t)\,v\bigr)\le r, so t u+(1βˆ’t) v∈BΛ‰dE(x,r)t\,u+(1-t)\,v\in\bar{B}_{d_E}(x,r). As uu, vv and tt were arbitrary, the closed ball is convex.

Claim 2. By claim 3 of Elementary Properties of the Closed Ball in a Metric Space, BˉdE(x,r)\bar{B}_{d_E}(x,r) is a closed subset of (Rn,TdE)(\mathbb{R}^n,\mathcal{T}_{d_E}), so by The Closure is the Smallest Closed Superset its closure in that topological space is BˉdE(x,r)\bar{B}_{d_E}(x,r) itself. By claim 2 of Elementary Properties of the Closed Ball in a Metric Space it is bounded in (Rn,dE)(\mathbb{R}^n,d_E). Hence The Closure of a Bounded Subset of Rn\mathbb{R}^n is Compact, applied to this bounded set, shows that BˉdE(x,r)\bar{B}_{d_E}(x,r) is compact in (Rn,TdE)(\mathbb{R}^n,\mathcal{T}_{d_E}).

Please log in to copy this version.

Citations

Loading…

Dependency Graph

0 prerequisites

Prerequisites

Loading...

Comments

Loading…