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ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28
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An additive f:RRf : \mathbb{R} \to \mathbb{R} that is not xcxx \mapsto 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\mathbb{R}^{2}, and every nonempty level set is dense in R\mathbb{R}

Example

Assume the Axiom of Choice (The Axiom of Choice), which enters through Assuming the Axiom of Choice, R\mathbb{R} has a Hamel basis over Q\mathbb{Q}: there is BRB \subseteq \mathbb{R} such that every real is a finite Q\mathbb{Q}-linear combination of elements of BB in exactly one way, and each basis vector carries a well-defined Q\mathbb{Q}-linear coefficient map and hence through Zorn's lemma. Fix a Hamel basis BB of R\mathbb{R} over the canonical copy QR\mathbb{Q} \subseteq \mathbb{R} of the rationals (The rationals embed densely in the reals, A field is a vector space over itself, and over any subfield KFK \subseteq F every FF-vector space is a KK-vector space by restricting the scalars, Vector space over a field), fix bBb_{\star} \in B, and let

f  :=  Λb:RRf \;:=\; \Lambda_{b_{\star}} : \mathbb{R} \to \mathbb{R}

be the coefficient map of bb_{\star} (Assuming the Axiom of Choice, R\mathbb{R} has a Hamel basis over Q\mathbb{Q}: there is BRB \subseteq \mathbb{R} such that every real is a finite Q\mathbb{Q}-linear combination of elements of BB in exactly one way, and each basis vector carries a well-defined Q\mathbb{Q}-linear coefficient map, claim 4). Write W:=Wb=span(B{b})W := W_{b_{\star}} = \operatorname{span}(B \setminus \{b_{\star}\}) (Linear combination of a finite list, and the span span(S)\operatorname{span}(S) as the smallest linear subspace containing SS). Then:

  1. ff is additive (Cauchy's functional equation f(x+y)=f(x)+f(y)f(x+y) = f(x) + f(y), and the additive functions RR\mathbb{R} \to \mathbb{R}) and is not of the form xcxx \mapsto cx for any real cc (FALSE: every additive f:RRf : \mathbb{R} \to \mathbb{R} is of the form xcxx \mapsto cx for a single real cc);
  2. ff is bounded neither above nor below on any nondegenerate interval (Lower bound, bounded below, bounded set, Intervals of R\mathbb{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\mathbb{R}, with the dictionary to monotone sequences), is of constant sign on none, and is continuous at no point of R\mathbb{R} (Continuity of f:ARf : A \to \mathbb{R} at a point of AA and on AA: the ε\varepsilon-δ\delta condition, its agreement with limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) at a limit point, and continuity at an isolated point);
  3. the graph {(x,f(x)):xR}\{(x,f(x)) : x \in \mathbb{R}\} is dense in R2\mathbb{R}^{2} for the metric dd_\infty (Rn\mathbb{R}^n as the set of functions nRn \to \mathbb{R}, and d1d_1, d2d_2, dd_\infty are metrics on it, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space);
  4. the values of ff are exactly the rationals, and for every rational rr the level set f1({r})={xR:f(x)=r}f^{-1}(\{r\}) = \{\, x \in \mathbb{R} : f(x) = r \,\} is dense in R\mathbb{R}; for an irrational vv the level set f1({v})f^{-1}(\{v\}) is empty.

Claim 2 is the contrapositive of Six regularity conditions each force an additive f:RRf : \mathbb{R} \to \mathbb{R} to be xf(1)xx \mapsto 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\mathbb{R}^{2} 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 BB of R\mathbb{R} over Q\mathbb{Q}; a fixed bBb_{\star} \in B; the coefficient map f=Λbf = \Lambda_{b_{\star}} and W=span(B{b})W = \operatorname{span}(B \setminus \{b_{\star}\}).

[A1]

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

[L1]

Assume the Axiom of Choice. Then a Hamel basis BB exists; for bBb_{\star} \in B the coefficient map Λb:RQ\Lambda_{b_{\star}} : \mathbb{R} \to \mathbb{Q} is well defined, additive, Q\mathbb{Q}-homogeneous, has range all of Q\mathbb{Q}, has {x:Λb(x)=0}=W\{x : \Lambda_{b_{\star}}(x) = 0\} = W, and W{0}W \ne \{0\} (Assuming the Axiom of Choice, R\mathbb{R} has a Hamel basis over Q\mathbb{Q}: there is BRB \subseteq \mathbb{R} such that every real is a finite Q\mathbb{Q}-linear combination of elements of BB in exactly one way, and each basis vector carries a well-defined Q\mathbb{Q}-linear coefficient map, claims 1, 4 and 5, Linear combination of a finite list, and the span span(S)\operatorname{span}(S) as the smallest linear subspace containing SS, Linear subspace of a vector space).

[L2]

There is an additive RR\mathbb{R} \to \mathbb{R} that is not of the form xcxx \mapsto cx, namely a coefficient map Λb\Lambda_{b_{\star}}: it takes only rational values while c0c \ne 0 would force irrational values (FALSE: every additive f:RRf : \mathbb{R} \to \mathbb{R} is of the form xcxx \mapsto cx for a single real cc, Both Q\mathbb{Q} and RQ\mathbb{R} \setminus \mathbb{Q} are dense in R\mathbb{R}, and every nonempty open subset of R\mathbb{R} is uncountable).

[L3]

If an additive g:RRg : \mathbb{R} \to \mathbb{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\mathbb{R}^{2}, then g(x)=g(1)xg(x) = g(1)x for every real xx (Six regularity conditions each force an additive f:RRf : \mathbb{R} \to \mathbb{R} to be xf(1)xx \mapsto 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\mathbb{R}^{2}).

[L5]

WW is a linear subspace of R\mathbb{R} over Q\mathbb{Q}, so wWw \in W and qQq \in \mathbb{Q} give qwWqw \in W, and WW is closed under addition (Linear subspace of a vector space, Linear combination of a finite list, and the span span(S)\operatorname{span}(S) as the smallest linear subspace containing SS).

[L6]

Strictly between any two distinct reals there lies a rational, and R\mathbb{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 BB and bBb_{\star} \in B, and put f:=Λbf := \Lambda_{b_{\star}} and W:=WbW := W_{b_{\star}}.

A1L1construct
2.1

Claim 1: ff is additive, and it is not of the form xcxx \mapsto cx for any real cc.

step 1.1L1L2
2.2

Claim 4, the range: the range of ff is exactly Q\mathbb{Q}, so f1({v})=f^{-1}(\{v\}) = \varnothing for every irrational vv and f1({r})f^{-1}(\{r\}) \ne \varnothing for every rational rr.

step 1.1L1
2.3

WW is dense in R\mathbb{R}: by [L1] there is w0Ww_{0} \in W with w00w_{0} \ne 0, and qw0Wq w_{0} \in W for every rational qq; given reals u<vu < v, the two reals u/w0u/w_{0} and v/w0v/w_{0} are distinct, so a rational qq lies strictly between them, and then qw0q w_{0} lies strictly between uu and vv if w0>0w_{0} > 0, and strictly between vv and uu if w0<0w_{0} < 0. Either way WW meets (u,v)(u,v).

step 1.1L1L5L6
3.1

Claim 2, clause by clause. Were ff 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)xf(x) = f(1)x for every real xx, contradicting step 2.1. So none of the five holds.

step 2.1L3
3.2

Claim 3: were the graph of ff not dense in R2\mathbb{R}^{2}, 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 rr the level set f1({r})f^{-1}(\{r\}) is xr+Wx_{r} + W for any xrx_{r} with f(xr)=rf(x_{r}) = r: indeed f(y)=rf(y) = r holds exactly when f(yxr)=f(y)f(xr)=0f(y - x_{r}) = f(y) - f(x_{r}) = 0, that is exactly when yxrWy - x_{r} \in W. Here f(x)=f(x)f(-x) = -f(x) follows from additivity.

step 1.1step 2.2L1L7
4.1

Each such level set is dense in R\mathbb{R}: given reals u<vu < v, the interval (uxr, vxr)(u - x_{r},\ v - x_{r}) meets WW by step 2.3, say in ww, and then xr+wf1({r})x_{r} + w \in f^{-1}(\{r\}) lies in (u,v)(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

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