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ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28
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An additive f:R→R that is not x↦cx: the coefficient of one fixed Hamel basis vector. It is unbounded above and below on every nondegenerate interval, its graph is dense in R2, and every nonempty level set is dense in R

Example

Assume the Axiom of Choice (The Axiom of Choice), which enters through Assuming the Axiom of Choice, R has a Hamel basis over Q: there is B⊆R such that every real is a finite Q-linear combination of elements of B in exactly one way, and each basis vector carries a well-defined Q-linear coefficient map and hence through Zorn's lemma. Fix a Hamel basis B of R over the canonical copy Q⊆R of the rationals (The rationals embed densely in the reals, A field is a vector space over itself, and over any subfield K⊆F every F-vector space is a K-vector space by restricting the scalars, Vector space over a field), fix b⋆∈B, and let

f  :=  Λb⋆:R→R

be the coefficient map of b⋆ (Assuming the Axiom of Choice, R has a Hamel basis over Q: there is B⊆R such that every real is a finite Q-linear combination of elements of B in exactly one way, and each basis vector carries a well-defined Q-linear coefficient map, claim 4). Write W:=Wb⋆=span⁡(B∖{b⋆}) (Linear combination of a finite list, and the span span⁡(S) as the smallest linear subspace containing S). Then:

  1. f is additive (Cauchy's functional equation f(x+y)=f(x)+f(y), and the additive functions R→R) and is not of the form x↦cx for any real c (FALSE: every additive f:R→R is of the form x↦cx for a single real c);
  2. f is bounded neither above nor below on any nondegenerate interval (Lower bound, bounded below, bounded set, Intervals of R: the nine order-convex forms, nondegeneracy, and length), is monotone on no nondegenerate interval (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R, with the dictionary to monotone sequences), is of constant sign on none, and is continuous at no point of R (Continuity of f:A→R at a point of A and on A: the ε-δ condition, its agreement with lim⁡x→cf(x)=f(c) at a limit point, and continuity at an isolated point);
  3. the graph {(x,f(x)):x∈R} is dense in R2 for the metric d∞ (Rn as the set of functions n→R, and d1, d2, d∞ are metrics on it, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space);
  4. the values of f are exactly the rationals, and for every rational r the level set f−1({r})={ x∈R:f(x)=r } is dense in R; for an irrational v the level set f−1({v}) is empty.

Claim 2 is the contrapositive of Six regularity conditions each force an additive f:R→R to be x↦f(1)x: continuity at a single point, monotonicity on a nondegenerate interval, boundedness above on one, boundedness below on one, constancy of sign on one, and a graph that is not dense in R2 applied to claim 1, clause by clause, and claim 3 is the contrapositive of its sixth clause.

Facts & Assumptions

Given: The Axiom of Choice; a Hamel basis B of R over Q; a fixed b⋆∈B; the coefficient map f=Λb⋆ and W=span⁡(B∖{b⋆}).

[A1]

The Axiom of Choice (The Axiom of Choice, Zorn's lemma).

[L1]

Assume the Axiom of Choice. Then a Hamel basis B exists; for b⋆∈B the coefficient map Λb⋆:R→Q is well defined, additive, Q-homogeneous, has range all of Q, has {x:Λb⋆(x)=0}=W, and W≠{0} (Assuming the Axiom of Choice, R has a Hamel basis over Q: there is B⊆R such that every real is a finite Q-linear combination of elements of B in exactly one way, and each basis vector carries a well-defined Q-linear coefficient map, claims 1, 4 and 5, Linear combination of a finite list, and the span span⁡(S) as the smallest linear subspace containing S, Linear subspace of a vector space).

[L2]

There is an additive R→R that is not of the form x↦cx, namely a coefficient map Λb⋆: it takes only rational values while c≠0 would force irrational values (FALSE: every additive f:R→R is of the form x↦cx for a single real c, Both Q and R∖Q are dense in R, and every nonempty open subset of R is uncountable).

[L3]

If an additive g:R→R is bounded above on a nondegenerate interval, or bounded below on one, or monotone on one, or of constant sign on one, or continuous at a single point, or has non-dense graph in R2, then g(x)=g(1)x for every real x (Six regularity conditions each force an additive f:R→R to be x↦f(1)x: continuity at a single point, monotonicity on a nondegenerate interval, boundedness above on one, boundedness below on one, constancy of sign on one, and a graph that is not dense in R2).

[L5]

W is a linear subspace of R over Q, so w∈W and q∈Q give qw∈W, and W is closed under addition (Linear subspace of a vector space, Linear combination of a finite list, and the span span⁡(S) as the smallest linear subspace containing S).

[L6]

Strictly between any two distinct reals there lies a rational, and R is an ordered field (The rationals embed densely in the reals, Complete ordered field (least-upper-bound property)).

Verification

technique · constructive
1.1

Assume the Axiom of Choice, fix B and b⋆∈B, and put f:=Λb⋆ and W:=Wb⋆.

A1L1construct
2.1

Claim 1: f is additive, and it is not of the form x↦cx for any real c.

step 1.1L1L2
2.2

Claim 4, the range: the range of f is exactly Q, so f−1({v})=∅ for every irrational v and f−1({r})≠∅ for every rational r.

step 1.1L1
2.3

W is dense in R: by [L1] there is w0∈W with w0≠0, and qw0∈W for every rational q; given reals u<v, the two reals u/w0 and v/w0 are distinct, so a rational q lies strictly between them, and then qw0 lies strictly between u and v if w0>0, and strictly between v and u if w0<0. Either way W meets (u,v).

step 1.1L1L5L6
3.1

Claim 2, clause by clause. Were f bounded above on a nondegenerate interval, or bounded below on one, or monotone on one, or of constant sign on one, or continuous at a single point, the regularity theorem would give f(x)=f(1)x for every real x, contradicting step 2.1. So none of the five holds.

step 2.1L3
3.2

Claim 3: were the graph of f not dense in R2, the sixth clause of the regularity theorem would give the same contradiction. So the graph is dense.

step 2.1L3L4
3.3

For a rational r the level set f−1({r}) is xr+W for any xr with f(xr)=r: indeed f(y)=r holds exactly when f(y−xr)=f(y)−f(xr)=0, that is exactly when y−xr∈W. Here f(−x)=−f(x) follows from additivity.

step 1.1step 2.2L1L7
4.1

Each such level set is dense in R: given reals u<v, the interval (u−xr, v−xr) meets W by step 2.3, say in w, and then xr+w∈f−1({r}) lies in (u,v). Claim 4 is proved, and with steps 2.1, 3.1 and 3.2 so are claims 1, 2 and 3.

step 2.1step 3.1step 3.2step 2.2step 2.3step 3.3discharge-construct∎

Remarks

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Sources