Alphabeta MathHow statement and proof provenance work
The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
- Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
- AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
- AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.
These labels describe origin, not correctness: citations and verification chips remain separate evidence.
Countably many concentric circles give an injective immersion that is not an embedding
Statement refuted
Every injective immersion is an embedding, and every immersed submanifold inherits the subspace topology of its image.
Facts & Assumptions
Given: The disjoint union of one unit circle and the circles of radii for , with the componentwise inclusion .
The two displayed claims are the false statements under discussion (An injective immersion need not be an embedding, The intrinsic topology of an immersed submanifold need not be the subspace topology).
Countable disjoint unions of fixed-dimensional smooth manifolds are smooth manifolds, and the circles here are smooth one-manifolds (Countable disjoint unions of fixed-dimensional smooth manifolds are smooth manifolds, A regular level set is an embedded submanifold, For a natural the function is differentiable everywhere with derivative ; for it is the constant , with derivative ; for a natural the function is differentiable at every with derivative ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products and quotients: , , , and when ).
Counterexample
By [L1], is a smooth manifold. The restriction of to each component is the ordinary circle inclusion, so is an injective immersion.
In the intrinsic topology of , each component is open. In the subspace topology on , the unit circle component is not open because every neighbourhood of one of its points meets infinitely many outer circles. So is not a homeomorphism and the topologies do not agree.
Therefore this single example simultaneously refutes both claims in [F1].
Depends on
- An injective immersion need not be an embedding
- The intrinsic topology of an immersed submanifold need not be the subspace topology
- Countable disjoint unions of fixed-dimensional smooth manifolds are smooth manifolds
- A regular level set is an embedded submanifold
- For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$; for $n = 0$ it is the constant $1$, with derivative $0$; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term
- Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$, $(\alpha f)'(c) = \alpha f'(c)$, $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$, and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
35 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
- John M. Lee, Introduction to Smooth Manifolds, Embeddings (standard reference, not scraped)