trureturing — Discovering truth

bibkey: hieu2026connectedness authors: Vu Trung Hieu year: 2026 title: “A connectedness result for convex polynomial multi-objective optimization without boundedness” doi: null url: https://arxiv.org/abs/2610.01029v1 claim: “Section 4.3 asks whether the weak Pareto solution set remains connected for general continuous convex data on a closed convex feasible set; Theorem 1 proves connectedness for convex polynomial data without boundedness or constraint qualification.” strata_touched:

• D5/S3/Analytic/Convexity/WeakParetoDisconnected license: citation-only triage: anchor

Locator

Author text: https://arxiv.org/abs/2610.01029v1. Vu Trung Hieu, A connectedness result for convex polynomial multi-objective optimization without boundedness, arXiv:2610.01029v1 (2026), §4.3 “Scope of the hypotheses”.

The paper asks whether ${\mathrm{Sol}}^w(K,f)$ is connected when $K$ is closed and convex and the objective functions are convex and continuous without the polynomial assumption. It states that no explicit example of a disconnected weak Pareto solution set is known, and that Theorem 1 rules out such an example for convex polynomial data.

Connectedness question

For the weak Pareto solution set

$${\mathrm{Sol}}^w(K,f)=\{z\in K:\nexists u\in K\text{ with }{f}_i(u)<f_i(z)\text{ for every }i\},$$

Theorem 1 gives connectedness for convex polynomial objectives on a closed convex feasible set without boundedness or a constraint qualification. The §4.3 question concerns continuous convex objectives in the general case.

Repository-derived counterexample. The companion Lean module uses $K=[0,\infty )\times [0,2]$ and $h(x)=\sqrt{x^2+1}-x$. Its objectives are $f_1(x,y)=y^2$ and $f_2(x,y)=(\max (h(x)-y/2,0){)}^2-y/2$.

The weak Pareto set has the proved boundary pieces

$$A=\{(x,0):x\ge 0\},B=\{(x,y):x\ge 0,2h(x)\le y\le 2\},$$

in the sense established by the two boundary lemmas: every point on either boundary is weak Pareto, and every weak Pareto point lies on one of them. A point with $0<y<2h(x)$ is strictly dominated by $(1/v,v)$ for a suitable $v>0$. Since $h(x)>0$, the open sets $\{y<h(x)\}$ and $\{h(x)<y\}$ separate the two nonempty parts of the weak Pareto set.

Motivation

The example answers the paper’s §4.3 question in the continuous convex setting. The analytic arguments below show that the two pieces approach one another as $x\to \infty$ and that $C^1$ regularity, semi-algebraicity, and attainment of every nonnegative scalarization do not restore connectedness.

Gap

The paper states that no explicit example of a disconnected weak Pareto solution set was known. Warburton’s connectedness theorem for bounded $K$ does not apply because this $K$ is unbounded. Helbig’s result for compact lower level sets does not apply either. The example therefore fills the stated general convex gap while retaining a closed convex feasible set and continuous convex objectives.

Route

Use the closed convex strip

$$K=\{(x,y)\in {\mathbb{R}}^2:0\le x,0\le y\le 2\}$$

and the function $h(x)=\sqrt{x^2+1}-x$. The identity $h(x)>0$ separates the axes and upper boundary $y=2h(x)$. Convexity of $h$ follows from the Euclidean norm representation $h(x)=\Vert (x,1)\Vert -x$; convexity of $f_1$ follows from Mathlib’s even-power convexity and affine composition, and convexity of $f_2$ follows from Mathlib’s supremum, nonnegative-power, and affine-subtraction results. Both objectives are continuous.

For a point in the open gap, write the second objective as $d$. Choose $v=(\max (0,-2d)+y)/2$; then $0<v<y$, and $(1/v,v)\in K$ strictly improves both objectives. The reciprocal estimate $h(1/v)<v/2$ makes the second objective strict. Thus only the two boundary pieces can be weak Pareto. The open-set separation then contradicts IsPreconnected.

Falsifier

The universal assertion, with $n,m\ge 1$, is falsified by $n=m=2$, the strip $K$ above, and the two displayed objectives. The module proves that $K$ is nonempty, closed, and convex, and that each objective is continuous and convex on the full Euclidean space. It proves non-preconnectedness of the resulting weak Pareto set, so no boundedness or constraint qualification is used in the counterexample.

Evidence

The public declarations are weakPareto, claim, and result; the instance data and supporting lemmas are private to D5/S3/Analytic/Convexity/WeakParetoDisconnected.lean.

The Lean proof establishes the instance properties, the two boundary inclusions (weak_of_boundary and weak_boundary), strict domination throughout the open gap (dominated_in_gap), and non-preconnectedness (not_preconnected). The theorem D5.S3.Analytic.Convexity.WeakParetoDisconnected.result proves ¬ claim.

The accepted axiom closure is $\{\mathrm{propext},Classical.choice,Quot.sound\}$; there is no sorry or additional axiom. The proof uses the square-root identities, Euclidean norm convexity, and exact real inequalities in Mathlib.

The module does not separately name an equality theorem for $W=A\cup B$; the two boundary lemmas together with the gap-domination lemma provide the corresponding characterization pieces used by the refutation.

Triage

What the settlement shows

1. Proved (Lean) — disconnected weak Pareto set. The displayed closed convex strip and continuous convex objectives satisfy the universal hypotheses, while their weak Pareto set is not preconnected.

2. Proved (Lean) — boundary characterization pieces. Every feasible point with $y=0$ or $y\ge 2h(x)$ is weak Pareto, and every weak Pareto point satisfies $y=0$ or $y\ge 2h(x)$. Every feasible point in the remaining positive gap has a strict dominator.

3. Proved (analytic argument, not in Lean) — asymptotic separation and regularity. Rationalization gives $h(x)=1/(\sqrt{x^2+1}+x)$ for $x\ge 0$, hence $h(x)\to 0$ as $x\to \infty$. The distance from $(x,2h(x))\in B$ to $(x,0)\in A$ is $2h(x)$, tending to zero, although the Lean open-set separation holds at every finite point. The function $h$ is smooth since $x^2+1>0$. The function $\varphi (t)=\max (t,0{)}^2$ is $C^1$: its derivative is $2\max (t,0)$, including derivative zero at $t=0$. Thus $f_1$ and $f_2=\varphi (h(x)-y/2)-y/2$ are $C^1$. The graph of $h$ is described by $t+x\ge 0$ and $(t+x{)}^2=x^2+1$; the two cases defining $\varphi$ are polynomial. Composition and polynomial operations preserve semi-algebraicity, so the objectives and strip are semi-algebraic. Neither $C^1$ regularity nor semi-algebraicity alone therefore guarantees connectedness in this setting.

4. Proved (analytic argument, not in Lean) — every scalarization attains a minimum. Let $\alpha ,\beta \ge 0$ be not both zero. For $\alpha ,\beta >0$, set

   $$y_{\ast }=\min \left(2,\frac{\beta }{4\alpha }\right)>0.$$

   On $K$, the nonnegative square term gives

   $$\alpha f_1(x,y)+\beta f_2(x,y)\ge \alpha y^2-\frac{\beta y}{2}\ge \alpha y_{\ast }^2-\frac{\beta y_{\ast }}{2}.$$

   The second inequality follows by completing the square, $\alpha y^2-\beta y/2=\alpha (y-\beta /(4\alpha ){)}^2-{\beta }^2/(16\alpha )$, and minimizing on $0\le y\le 2$. Since $h^{\prime }(x)=x/\sqrt{x^2+1}-1<0$ and $h(1/y_{\ast })<y_{\ast }/2$ (h_reciprocal), the positive-part term vanishes on the whole ray $x\ge 1/y_{\ast },y=y_{\ast }$. Every feasible point on that ray attains the displayed bound. More precisely, equality requires $y=y_{\ast }$ and $h(x)\le y_{\ast }/2$; squaring the latter inequality (both sides of $\sqrt{x^2+1}\le x+y_{\ast }/2$ are nonnegative) gives the full minimizer ray $x\ge 1/y_{\ast }-y_{\ast }/4,y=y_{\ast }$. Its endpoint is nonnegative since $y_{\ast }\le 2$.

   For $\alpha >0,\beta =0$, the minimum is zero, attained at $(0,0)$ and on the whole ray $x\ge 0,y=0$. For $\alpha =0,\beta >0$, the lower bound $f_2\ge -y/2\ge -1$ is attained at $(0,2)$ and on the whole ray $x\ge 0,y=2$, because $h(x)\le h(0)=1$. Equality forces $y=2$, so this is the full minimizer set. All nonzero nonnegative scalarizations thus have unbounded minimizer rays. Attainment of every scalarization does not imply connectedness of the weak Pareto set.

5. Open — neighboring question. The Pareto (non-weak) half of Yen’s Question 9.8 for convex polynomials is not settled by this delivery.

ASSUMED-UNVERIFIED

Zhong–Fan, Journal of Guangxi Normal University 39(3) (2021), 62–68, is reported in its abstract to prove connectedness on unbounded sets via upper semicontinuity of the scalarized solution map. Only the abstract was read; applicability to this instance remains unresolved until the actual hypotheses are checked. Scalarization attainment is established above, so non-attainment cannot explain any applicability boundary.

The asymptotic limit, $C^1$ regularity, semi-algebraicity, and scalarization attainment above are analytic arguments, not Lean results. The literature’s specific connectedness mechanisms are not verified by this delivery.