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CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
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g(x,y)=xy/(x2+y2)g(x,y) = xy/(x^{2}+y^{2}), extended by g(0,0)=0g(0,0)=0, is continuous in each variable separately and not continuous at the origin

Statement refuted

Refuted claim: a function g:R2Rg : \mathbb{R}^{2} \to \mathbb{R} that is continuous in each variable separately — that is, for which tg(t,b)t \mapsto g(t,b) and tg(a,t)t \mapsto g(a,t) are continuous on R\mathbb{R} for every fixed aa and bb (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) — is continuous as a map (R2,d2)(R,dR)(\mathbb{R}^{2}, d_2) \to (\mathbb{R}, d_{\mathbb{R}}) (Vector-valued functions f:ARmf : A \to \mathbb{R}^m, their limits and continuity, with the dictionary to the metric notions, Continuity of a map between metric spaces, at a point and globally, in the ε\varepsilon-δ\delta form, Rn\mathbb{R}^n as the set of functions nRn \to \mathbb{R}, and d1d_1, d2d_2, dd_\infty are metrics on it).

The witness. Define g:R2Rg : \mathbb{R}^{2} \to \mathbb{R} by

g(p)  :=  p0p1p02+p12  for p0,g(0):=0,g(p) \;:=\; \frac{p_0\,p_1}{p_0^{2}+p_1^{2}} \ \text{ for } p \ne 0, \qquad g(0) := 0 ,

writing p=(p0,p1)p = (p_0,p_1) for an element of R2\mathbb{R}^{2}, the set of functions 2R2 \to \mathbb{R} (Rn\mathbb{R}^n as the set of functions nRn \to \mathbb{R}, and d1d_1, d2d_2, dd_\infty are metrics on it). The quotient is defined for p0p \ne 0 because p02+p12=p22>0p_0^{2}+p_1^{2} = \lVert p\rVert_2^{2} > 0 there (The Euclidean inner product x,y=k<nxkyk\langle x,y\rangle = \sum_{k<n} x_k y_k on Rn\mathbb{R}^n, A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Then gg is continuous in each variable separately at every point, and gg is not continuous at 00.

This is the first function on R2\mathbb{R}^{2} whose continuity this library studies, and its domain is R2\mathbb{R}^{2} with the published metric d2d_2, not an informal plane.

Facts & Assumptions

Given: The function g:R2Rg : \mathbb{R}^{2} \to \mathbb{R} above; the sequence p(k):=(1/ι(k+1), 1/ι(k+1))p^{(k)} := \bigl(1/\iota(k+1),\ 1/\iota(k+1)\bigr) in R2\mathbb{R}^{2} (Sequences of reals: bounded, eventually, frequently, tails, subsequences, The canonical natural ι(n)=n1F\iota(n) = n \cdot 1_F of a field).

[A1]

The refuted claim, at this gg: separate continuity everywhere implies continuity as a map (R2,d2)(R,dR)(\mathbb{R}^{2},d_2) \to (\mathbb{R},d_{\mathbb{R}}).

Counterexample

technique · direct
1.1

For a fixed real b0b \ne 0 the function tg(t,b)=tb/(t2+b2)t \mapsto g(t,b) = tb/(t^{2}+b^{2}) is a quotient of two polynomial functions of tt whose denominator never vanishes, since t2+b2b2>0t^{2}+b^{2} \ge b^{2} > 0; so it is continuous on R\mathbb{R}.

L3L5
1.2

For b=0b = 0 the function tg(t,0)t \mapsto g(t,0) is constantly 00: at t0t \ne 0 its value is t0/(t2+0)=0t\cdot 0/(t^{2}+0) = 0, and at t=0t = 0 it is g(0)=0g(0) = 0. A constant function is continuous.

L3L5
1.3

Each coordinate sequence of (p(k))\bigl(p^{(k)}\bigr) is k1/ι(k+1)k \mapsto 1/\iota(k+1), which converges to 00: given a rational ε>0\varepsilon > 0, an index KK with 1/ι(K+1)<ε1/\iota(K+1) < \varepsilon gives 0<1/ι(k+1)1/ι(K+1)<ε0 < 1/\iota(k+1) \le 1/\iota(K+1) < \varepsilon for every kKk \ge K. Hence p(k)0p^{(k)} \to 0 in (R2,d2)(\mathbb{R}^{2},d_2).

L2L4
2.1

By the symmetry g(p0,p1)=g(p1,p0)g(p_0,p_1) = g(p_1,p_0), the same two arguments give continuity of tg(a,t)t \mapsto g(a,t) for every fixed real aa.

step 1.1step 1.2
2.2

For every kk the point p(k)p^{(k)} is nonzero, and with u:=1/ι(k+1)u := 1/\iota(k+1) its value is g(p(k))=uu/(u2+u2)=u2/(ι(2)u2)=1/ι(2)g(p^{(k)}) = u\cdot u/(u^{2}+u^{2}) = u^{2}/\bigl(\iota(2)u^{2}\bigr) = 1/\iota(2).

step 1.3L4L5
3.1

So gg is continuous in each variable separately at every point of R2\mathbb{R}^{2}.

step 1.1step 1.2step 2.1
3.2

So the constant sequence (g(p(k)))\bigl(g(p^{(k)})\bigr) converges to 1/ι(2)1/\iota(2), while g(0)=0g(0) = 0 and 1/ι(2)01/\iota(2) \ne 0 because ι(2)>0\iota(2) > 0.

step 2.2L4
4.1

By the sequential characterisation of continuity, gg is not continuous at 00: the sequence p(k)0p^{(k)} \to 0 has g(p(k))↛g(0)g(p^{(k)}) \not\to g(0).

step 1.3step 2.2step 3.2L1
5.1

Steps 3.1 and 4.1 together refute [A1]: gg is separately continuous everywhere and is not continuous at the origin.

step 3.1step 4.1A1

Remarks

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