CounterexampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-09-10 (Codex)

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$g (x, y) = x y / (x^{2} + y^{2})$ , extended by $g (0, 0) = 0$ , is continuous in each variable separately and not continuous at the origin

Statement refuted

Refuted claim: a function $g : R^{2} \to R$ that is continuous in each variable separately — that is, for which $t \mapsto g (t, b)$ and $t \mapsto g (a, t)$ are continuous on $R$ for every fixed $a$ and $b$ (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point) — is continuous as a map $(R^{2}, d_{2}) \to (R, d_{R})$ (Vector-valued functions $f : A \to R^{m}$ , 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 $ε$ - $δ$ form, $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it).

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

$g (p) : = \frac{p_{0} p_{1}}{p_{0}^{2} + p_{1}^{2}} for p \neq 0, g (0) : = 0,$

writing $p = (p_{0}, p_{1})$ for an element of $R^{2}$ , the set of functions $2 \to R$ ( $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it). The quotient is defined for $p \neq 0$ because $p_{0}^{2} + p_{1}^{2} = ∥ p ∥_{2}^{2} > 0$ there (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Then $g$ is continuous in each variable separately at every point, and $g$ is not continuous at $0$ .

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

Facts & Assumptions

Given: The function $g : R^{2} \to R$ above; the sequence $p^{(k)} : = (1 / ι (k + 1), 1 / ι (k + 1))$ in $R^{2}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[A1]

The refuted claim, at this $g$ : separate continuity everywhere implies continuity as a map $(R^{2}, d_{2}) \to (R, d_{R})$ .

[L1]

A real-valued function on a metric space that is continuous at $q$ preserves every sequence converging to $q$ : if $p^{(k)} \to q$ , then $g (p^{(k)}) \to g (q)$ (Vector-valued functions $f : A \to R^{m}$ , 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 $ε$ - $δ$ form, Metric continuity characterisations, with countable choice for the sequential converse proof step 1.5, Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric, The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded).

For completeness, the pointwise implication in [L1] follows directly from the definitions: given a real $ε > 0$ , continuity at $q$ supplies a real $δ > 0$ such that $d (p, q) < δ$ implies $∣ g (p) - g (q) ∣ < ε$ . Convergence $p^{(k)} \to q$ supplies $K$ with $d (p^{(k)}, q) < δ$ for all $k \geq K$ , using the rational-to-real tolerance agreement in Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ . Hence $∣ g (p^{(k)}) - g (q) ∣ < ε$ for all $k \geq K$ . This proves the pointwise implication without assuming continuity elsewhere or any choice principle.

[L2]

Convergence in $(R^{2}, d_{2})$ is componentwise (For $n \geq 1$ a sequence in $R^{n}$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $R^{n}$ is complete in every norm clause 1, Convergence of a sequence in a metric space: $x_{k} \to x$ iff $d (x_{k}, x) \to 0$ in $R$ , Each $∥ \cdot ∥_{p}$ is a norm on $R^{n}$ , and the induced metrics are exactly $d_{1}$ , $d_{2}$ and $d_{\infty}$ of the published metric-spaces page, The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ , The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L3]

Algebra of continuous real functions on a subset of $R$ : sums, products and quotients with nonvanishing denominator of continuous functions are continuous, and every polynomial function is continuous (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, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, Dictionary: for $A \subseteq R$ with the metric $d (x, y) = ∣ x - y ∣$ , continuity and uniform continuity of $f : A \to R$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $R$ is compact in the open-cover sense of $R$ exactly when it is a compact metric subspace).

[L4]

The canonical natural: $ι (k + 1) > 0$ for every $k \in N$ , and for every real $ε > 0$ there is $k$ with $1 / ι (k + 1) < ε$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Inverses of positives are positive, and reciprocation reverses order).

[L5]

$∥ p ∥_{2}^{2} = p_{0}^{2} + p_{1}^{2}$ , which is $0$ only for $p = 0$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ , Integer powers $a^{m}$ , 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).

Counterexample

technique · direct

1.1

For a fixed real $b \neq 0$ the function $t \mapsto g (t, b) = t b / (t^{2} + b^{2})$ is a quotient of two polynomial functions of $t$ whose denominator never vanishes, since $t^{2} + b^{2} \geq b^{2} > 0$ ; so it is continuous on $R$ .

L3L5

1.2

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

L3L5

1.3

Each coordinate sequence of $(p^{(k)})$ is $k \mapsto 1 / ι (k + 1)$ , which converges to $0$ : given a rational $ε > 0$ , an index $K$ with $1 / ι (K + 1) < ε$ gives $0 < 1 / ι (k + 1) \leq 1 / ι (K + 1) < ε$ for every $k \geq K$ . Hence $p^{(k)} \to 0$ in $(R^{2}, d_{2})$ .

L2L4

2.1

By the symmetry $g (p_{0}, p_{1}) = g (p_{1}, p_{0})$ , the same two arguments give continuity of $t \mapsto g (a, t)$ for every fixed real $a$ .

step 1.1step 1.2

2.2

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

step 1.3L4L5

3.1

So $g$ is continuous in each variable separately at every point of $R^{2}$ .

step 1.1step 1.2step 2.1

3.2

So the constant sequence $(g (p^{(k)}))$ converges to $1 / ι (2)$ , while $g (0) = 0$ and $1 / ι (2) \neq 0$ because $ι (2) > 0$ . More directly, this sequence cannot converge to $g (0) = 0$ : its distance from $0$ is always $1 / ι (2)$ , so the convergence test fails at the positive rational tolerance $1 / ι (4)$ .

step 2.2L4

4.1

By the contrapositive of the fact that continuity preserves convergent sequences, $g$ is not continuous at $0$ : the sequence $p^{(k)} \to 0$ has $g (p^{(k)}) \to̸ g (0)$ .

step 1.3step 2.2step 3.2L1

5.1

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

step 3.1step 4.1A1∎

Remarks

What the sequence sees. Along the line $p_{1} = p_{0}$ the value of $g$ is constantly $1 / ι (2)$ off the origin, and points of that line come arbitrarily close to the origin; along either axis the value is constantly $0$ . So the two partial functions through the origin cannot detect what a general approach does, and that is the whole phenomenon.
Separate continuity is strictly weaker, and no repair is proposed here. What the refuted claim would need is a hypothesis controlling the two variables together — joint continuity is exactly such a hypothesis, and it is what Vector-valued functions $f : A \to R^{m}$ , their limits and continuity, with the dictionary to the metric notions defines. Nothing here claims that any weaker hypothesis suffices.
Nothing is claimed about $g$ away from the origin. The refutation needs only the behaviour of $g$ at $0$ together with the two partial functions, and that is all that is proved. In particular this item does not assert that $g$ is continuous at the points $p \neq 0$ , true though that is; establishing it would need an algebra of continuous real-valued functions on a metric domain, which 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 provides for sums, scalar multiples and inner products but not for quotients.

Depends on

Used by

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Sources

Continuous function (Wikipedia) (standard reference, not scraped)
Multivariable calculus (Wikipedia) (standard reference, not scraped)
Harvard Math 21a, separately continuous but not jointly continuous example (standard reference, not scraped)

g(x,y)=xy/(x2+y2)g(x,y)=xy/(x2+y2), extended by g(0,0)=0g(0,0)=0, is continuous in each variable separately and not continuous at the origin