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CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-31
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For n1n\ge1, every Euclidean closed ball and every Euclidean sphere of positive radius is compact

Statement

For n1n\ge1, cRnc\in\mathbb{R}^n, and r>0r>0, the Euclidean closed ball B2(c,r)\overline B_2(c,r) and Euclidean sphere S2(c,r)S_2(c,r) are compact.

Facts & Assumptions

Given: n1n\ge1, cRnc\in\mathbb{R}^n, and r>0r>0.

[L1]

The sets B2(c,r)\overline B_2(c,r) and S2(c,r)S_2(c,r) are respectively the points satisfying xc2r\lVert x-c\rVert_2\le r and xc2=r\lVert x-c\rVert_2=r (Euclidean spheres and closed balls as subspaces of Rn\mathbb{R}^n).

[L2]

The Euclidean norm is continuous and satisfies u2v2uv2|\lVert u\rVert_2-\lVert v\rVert_2|\le\lVert u-v\rVert_2 (The finite and reverse triangle inequalities for a norm; and for n1n \ge 1 every norm NN on Rn\mathbb{R}^n satisfies N(x)Cx1N(x) \le C\lVert x\rVert_1 and is Lipschitz, hence continuous, for d2d_2).

Proof

technique · direct
1.1

Both sets are bounded: B2(c,r)\overline B_2(c,r) lies in the ball of radius r+1r+1 about cc, and S2(c,r)B2(c,r)S_2(c,r)\subseteq\overline B_2(c,r).

L1L4
1.2

The complement of B2(c,r)\overline B_2(c,r) is open: if xc2>r\lVert x-c\rVert_2>r, then the ball about xx of radius (xc2r)/2(\lVert x-c\rVert_2-r)/2 stays in the complement by [L2].

L1L2L4
1.3

The complement of S2(c,r)S_2(c,r) is open: if xc2r\lVert x-c\rVert_2\ne r, then a ball about xx of radius xc2r/2|\lVert x-c\rVert_2-r|/2 avoids the sphere by [L2].

L1L2L4
2.1

Thus both sets are closed and bounded, hence compact by [L3].

step 1.1step 1.2step 1.3L3

Depends on

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