TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-06 (claude-sonnet-5)

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For $n \geq 1$ , the map $H (x, t) = ((1 - t) + t / ∥ x ∥_{2}) x$ is continuous on $(R^{n} ∖ {0}) \times [0, 1]$ , starts at $x$ , ends at radial normalisation, fixes the unit sphere, and never reaches $0$

Statement

For $n \geq 1$ , put $P = R^{n} ∖ {0}$ and define

$H : P \times [0, 1] \to P, H (x, t) = ((1 - t) + t / ∥ x ∥_{2}) x .$

Then $H$ is continuous, $H (x, 0) = x$ , $H (x, 1) = x / ∥ x ∥_{2}$ , $H (s, t) = s$ for $s \in S^{n - 1}$ , and $H (x, t) \neq 0$ .

Facts & Assumptions

Given: $x \in P$ and $t \in [0, 1]$ .

[L1]

Radial normalisation is continuous on $P$ (Radial normalisation $x \mapsto x / ∥ x ∥_{2}$ is continuous on $R^{n} ∖ {0}$ ).

[L2]

Coordinate projections and the map into a product are continuous as stated by the product universal property (A map into a product is continuous iff each of its components is; the projections are continuous and open; and each projection is surjective when every factor is nonempty, which for an infinite index set uses the Axiom of Choice, The product set $\prod_{i \in I} X_{i}$ of functions choosing a point in each factor, the projections, the box topology, and the product topology as the initial topology of the projections; the empty product is a one-point space).

[L3]

The Euclidean norm is continuous for the Euclidean metric and is positive away from $0$ , and the unit sphere consists of its norm-one points (The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ , Euclidean spheres and closed balls as subspaces of $R^{n}$ ).

[L4]

For maps on a subset of a metric space, sums, scalar multiples and pointwise products of continuous real-valued functions are continuous, and a vector-valued map is continuous exactly when its coordinate functions are continuous (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions, clauses 1 and 3, the product being the case $m = 1$ of the inner product); composites of continuous maps are continuous (Continuity may be checked on any open cover, and on any finite closed cover; composites of continuous maps are continuous); and $u \mapsto 1 / u$ is continuous on $R ∖ {0}$ , this being clause 4 of (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function) with $A = R$ , numerator $1$ and denominator the identity.

[L5]

A map into a subspace is continuous exactly when its composite with the ambient inclusion is continuous (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace).

Proof

technique · direct

1.1L3

The scalar $c (x, t) : = (1 - t) + t / ∥ x ∥_{2}$ is positive, since $1 - t \geq 0$ , $t \geq 0$ , and $∥ x ∥_{2} > 0$ .

1.2L1L2L3L4

The coordinate projections on $P \times [0, 1]$ are continuous by [L2]. By [L3] the map $x \mapsto ∥ x ∥_{2}$ is continuous on $P$ and never $0$ there, so composing it with $u \mapsto 1 / u$ gives a continuous $x \mapsto 1 / ∥ x ∥_{2}$ by [L4]; hence $c (x, t) = (1 - t) + t / ∥ x ∥_{2}$ is continuous, being built from continuous functions by sums, scalar multiples and products as in [L4], and each coordinate $H_{i} (x, t) = c (x, t) x_{i}$ is continuous. Componentwise continuity in [L4] makes $H$ continuous as a map into $R^{n}$ .

1.3L3

Substituting $t = 0$ and $t = 1$ gives $H (x, 0) = x$ and $H (x, 1) = x / ∥ x ∥_{2}$ . If $s \in S^{n - 1}$ , then $∥ s ∥_{2} = 1$ and $H (s, t) = s$ .

2.1step 1.1step 1.2L5

Since $c (x, t) > 0$ and $x \neq 0$ , $H (x, t) \neq 0$ . Hence $H$ takes values in $P$ , and [L5] makes it continuous as a map $P \times [0, 1] \to P$ .

3.1step 2.1step 1.3∎

These identities and step 2.1 prove the statement.

Depends on

Dependency tree · 2 levels

Sources

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