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📝 Open and closed sets in multivariable calculus (13 MCQs)

📖 From Calculus • 14. Partial Derivatives Calculus • 13 questions available

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What is Open and closed sets in multivariable calculus?

Definition:
An open set contains none of its boundary points, while a closed set contains all its boundary points; these topological properties define domains for theorems.

Example:
The disk $x^2 + y^2 < 1$ is open, whereas $x^2 + y^2 \leq 1$ is closed and bounded (compact).

Reason:
Key results like the Extreme Value Theorem require closed bounded domains, while differentiability definitions typically assume open domains to allow approach from all directions.

Easy

Medium

Hard

📝 All Open and closed sets in multivariable calculus MCQs

Q1. A student claims that the domain $D = \{(x,y) \in \mathbb{R}^2 \mid x^2 + y^2 < 4, y \neq 0\}$ is an open set because it is defined by a strict inequality. Which statement best identifies the flaw in this reasoning?

A.The set is actually closed because removing a line creates boundary points.

B.Strict inequalities always define open sets, so the student is correct.

C.The set is neither open nor closed because removing

y=0

introduces boundary points within the interior region. ✅

D.The set is open but not connected, which invalidates the definition.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This question targets error analysis regarding topological definitions. While $x^2+y^2<4$ defines an open disk, removing the line $y=0$ means points on the x-axis are boundary points of $D$ that are not contained in $D$ . Since an open set must contain none of its boundary points, $D$ fails this criterion despite the strict inequality, illustrating that subset operations can destroy openness.

Q2. Consider the function $f(x,y) = \ln(x^2 - y)$ . When modeling a physical system where continuity is required for stability, which topological property best describes the maximal domain of definition for this function?

A.It is a closed set because the logarithm requires non-negative arguments.

B.It is an open set defined by the strict inequality

x^2 - y > 0

. ✅

C.It is a compact set bounded by the parabola

y = x^2

D.It is neither open nor closed due to the asymptotic behavior at infinity.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This application question connects function domains to topology. The natural logarithm requires a strictly positive argument, leading to the condition $x^2 - y > 0$ . The preimage of an open interval under a continuous function is open. Students must recognize that strict inequalities involving continuous functions typically generate open sets, distinguishing this from closed regions defined by non-strict inequalities or bounded compact regions.

Q3. Analyze the set $S = \{(x,y) \mid 0 < x \leq 1, 0 \leq y < 1\} \cup \{(0,0)\}$ in the context of partial derivative existence. Why is determining whether $S$ is open or closed critical before applying standard differentiability theorems?

A.Standard theorems require the domain to be both open and closed simultaneously.

B.Differentiability is only defined at interior points; since

S

contains boundary points like

(0,0)

, it is not open, limiting theorem applicability. ✅

C.The set is closed, so all partial derivatives automatically exist everywhere in

S

D.Topological properties are irrelevant to partial derivatives, which depend solely on algebraic limits.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This conceptual understanding question links topology to calculus prerequisites. Partial derivatives are inherently defined via limits requiring neighborhoods entirely within the domain. Since $S$ includes boundary points such as $(0,0)$ and segments where $x=1$ or $y=0$ , it lacks interior neighborhoods at those locations. Recognizing that $S$ is neither fully open nor closed prevents misapplication of theorems requiring open domains.

Q4. Given two subsets of $\mathbb{R}^2$ : $A = \{(x,y) \mid x^2 + y^2 \leq 1\}$ and $B = \{(x,y) \mid (x-1)^2 + y^2 < 1\}$ . If a optimization model requires the feasible region to be compact, which combination satisfies this requirement?

A.Only set

A

is compact because it is closed and bounded. ✅

B.Only set

B

is compact because open sets allow interior extrema.

C.Both

A

and

B

are compact as they are bounded subsets of

\mathbb{R}^2

D.Neither set is compact because

B

is open and

A

has a curved boundary.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This mixed concepts question integrates compactness with open/closed definitions. By the Heine-Borel theorem, a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded. Set $A$ includes its boundary ( $\leq$ ), making it closed and bounded, hence compact. Set $B$ uses a strict inequality ( $<$ ), making it open and thus not closed, failing the compactness criterion despite being bounded.

Q5. A contour plot shows level curves of $f(x,y)$ accumulating densely near the line $x=0$ , with no defined values for $x \leq 0$ . Based solely on this graphical evidence, what can be inferred about the topological nature of the domain $D$ ?

D

must be closed because the contours accumulate at a boundary.

D

is likely open on the right half-plane, as the accumulation suggests

x=0

is a boundary not included in

D

. ✅

D

is compact because the contours are bounded within the visible region.

D.The graph provides insufficient information to distinguish between open and closed sets.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This graph-based interpretation question requires translating visual density into topological properties. Dense contour accumulation near $x=0$ without defined values suggests $x=0$ acts as a boundary. If the domain extended to include $x=0$ , continuity would typically prevent such singular clustering unless specified. The absence of definition for $x \leq 0$ combined with approachable boundary behavior strongly indicates an open set like $\{(x,y) \mid x > 0\}$ , rather than a closed region including the axis.

Q6. In error analysis of a numerical PDE solver, the algorithm fails when grid points land exactly on $y = x^2$ . The programmer assumes the domain $\Omega = \{(x,y) \mid y > x^2\}$ is safe because it is open. What subtle topological issue might still cause failure near the parabolic boundary?

A.Open sets cannot be used for PDEs; only closed rectangles work.

B.Numerical discretization may place points outside

\Omega

due to floating-point errors, effectively sampling the boundary where the solution is undefined. ✅

C.The parabola

y=x^2

is not a smooth manifold, violating openness.

D.The set

\Omega

is actually closed because parabolas enclose regions.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This scenario-based error analysis highlights the gap between theoretical openness and computational reality. While $y > x^2$ is theoretically open, finite precision arithmetic means computed coordinates may satisfy $y \approx x^2$ or even $y < x^2$ due to rounding. Since the mathematical model is undefined on the boundary, these numerical artifacts cause failures. Understanding that practical implementation requires buffer zones beyond pure topological definitions is crucial for robust modeling.

Q7. Let $U \subset \mathbb{R}^2$ be an open set and $K \subset U$ be a compact subset. When proving that a continuously differentiable function $f: U \to \mathbb{R}$ attains a maximum on $K$ , which logical sequence correctly applies topological concepts?

A.Since

U

is open,

f

attains max on

U

; restrict to

K

B.Compactness of

K

ensures

f(K)

is compact in

\mathbb{R}

, hence closed and bounded, guaranteeing a maximum exists within

K

. ✅

C.Openness of

U

implies

K

is open, so extreme value theorem applies directly.

D.Maximum exists because partial derivatives vanish somewhere in

U

, regardless of

K

's topology.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: This multi-step reasoning question chains continuity, compactness, and extremal values. The Extreme Value Theorem requires a compact domain. Although $U$ is open (and thus not necessarily compact), the subset $K$ is explicitly compact. Continuous images of compact sets are compact in $\mathbb{R}$ , meaning $f(K)$ is closed and bounded. This guarantees attainment of supremum/infimum within $K$ , independent of $U$ 's openness, correcting the misconception that openness alone suffices.

Q8. A student argues that the intersection of infinitely many open sets in $\mathbb{R}^2$ must be open, citing that finite intersections preserve openness. Provide a counterexample demonstrating why this extension to infinite intersections fails in the context of partial derivatives domains.

A.The intersection of open balls

B_{1/n}(0,0)

for

n \in \mathbb{N}

yields the singleton

\{(0,0)\}

, which is closed. ✅

B.Infinite unions of closed sets are open, not intersections.

C.All infinite intersections of open sets remain open in metric spaces.

D.Partial derivatives cannot be defined on infinite intersections, making the question moot.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: This challenging Olympiad-style question tests deep understanding of topological axioms versus intuitive generalizations. While finite intersections of open sets are open, countable intersections need not be. The nested balls $B_{1/n}(0)$ shrink to the origin, producing a singleton set. Singletons are closed (complement is open) and not open in $\mathbb{R}^2$ . This distinction is vital when defining domains via limiting processes in advanced calculus, where assuming openness persists can invalidate derivative existence proofs.

Q9. When analyzing the set $E = \{(x,y) \in \mathbb{R}^2 \mid xy \geq 1\}$ , a learner incorrectly classifies it as open because the hyperbola $xy=1$ appears 'thin' in plots. Which rigorous criterion definitively proves $E$ is closed?

A.The complement

\{(x,y) \mid xy < 1\}

is open as the preimage of

(-\infty, 1)

under continuous

g(x,y)=xy

B.Hyperbolas always define open regions in multivariable calculus.

C.The set contains its boundary curve

xy=1

, satisfying the sequential characterization of closed sets.

D.Both A and C are valid rigorous proofs of closedness. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: This conceptual understanding question addresses visual misconceptions about 'thin' boundaries. Option A uses the topological definition via complements and continuity, while Option C uses the sequential closure property (containing all limit points). Both are mathematically equivalent and rigorous. The distractor exploits the false intuition that curves have zero area and thus don't affect openness. Emphasizing multiple valid proof methods strengthens conceptual flexibility beyond mere computation or visual estimation.

Q10. In a thermodynamic model, temperature $T(x,y)$ is defined only on an open region $\Omega$ . Engineers extend $T$ continuously to the boundary $\partial \Omega$ to compute heat flux. Why was the original domain required to be open for defining partial derivatives $T_x, T_y$ inside $\Omega$ ?

A.Partial derivatives require two-sided limits, which only exist at interior points of open sets. ✅

B.Boundary extensions make partial derivatives undefined everywhere.

C.Open sets ensure

T

is analytic, not just differentiable.

D.Thermodynamics prohibits closed domains due to entropy constraints.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This application question grounds abstract topology in physical modeling. Partial derivatives at a point require the function to be defined in a full neighborhood around that point to evaluate limits from all directions. Points in an open set possess such neighborhoods entirely within the domain. Boundary points lack this property, necessitating one-sided or directional derivatives instead. Understanding this motivates why models start with open domains before extending to boundaries for flux calculations, linking theory to engineering practice.

Q11. Compare the sets $A = \{(x,y) \mid x^2 + y^2 \leq 1\} \setminus \{(0,0)\}$ and $B = \{(x,y) \mid 0 < x^2 + y^2 \leq 1\}$ . In the context of singularity analysis for partial derivatives, how do their topological classifications differ?

A.Both are closed because they include the unit circle boundary.

B.Both are open because the origin is removed.

A

is neither open nor closed;

B

is also neither open nor closed, but they differ in boundary point inclusion. ✅

A

is closed and

B

is open.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: This comparative analysis question examines nuanced topological distinctions after point removal. Set $A$ removes an interior point from a closed disk, losing closedness (missing limit point) and never being open (contains boundary). Set $B$ excludes the origin but retains the outer boundary, also failing both definitions. Though both are neither open nor closed, their boundary structures differ: $A$ has an isolated missing interior point, while $B$ has a punctured interior with intact outer boundary. This affects singularity classification in PDEs.

Q12. A researcher models population density on $D = \{(x,y) \mid x > 0, y > 0, x+y < 10\}$ . They claim $\overline{D}$ (closure) represents the biologically feasible region including extinction states. What topological operation transforms $D$ into $\overline{D}$ , and why is this significant for boundary equilibrium analysis?

A.Taking the union with boundary

\partial D

adds limit points where density vanishes, enabling analysis of extinction equilibria. ✅

B.Removing interior points creates a closed set suitable for discrete models.

C.Intersecting with

\mathbb{R}^2_+

automatically closes the set.

D.Closure is irrelevant since populations are always positive.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: This scenario-based question applies closure operations to biological modeling. The open triangle $D$ excludes axes and hypotenuse where population components reach zero. Biological extinction corresponds precisely to these boundary points. Forming $\overline{D} = D \cup \partial D$ incorporates these limit points, creating a compact set where equilibrium analysis (including boundary equilibria) becomes possible via fixed-point theorems. This demonstrates how topological completion enables meaningful interpretation of edge cases in applied mathematics.

Q13. During peer review, a paper states: 'Since $f$ is continuous on the open set $U$ , it attains a minimum on $U$ .' As a reviewer identifying topological errors, which correction addresses the fundamental flaw?

A.Continuity on open sets guarantees minima only if

U

is also bounded.

B.Open sets lack guaranteed minima; the claim should specify a compact subset

K \subset U

or additional coercivity conditions. ✅

C.All continuous functions attain minima on any subset of

\mathbb{R}^n

D.The error is merely semantic; 'minimum' should be replaced with 'infimum'.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: This error analysis question targets a pervasive misconception in advanced calculus. Continuity alone on an open set does not guarantee extremum attainment; consider $f(x)=x$ on $(0,1)$ . The Extreme Value Theorem requires compactness. Valid corrections involve either restricting to compact subsets or imposing growth conditions (coercivity) ensuring sublevel sets are compact. Simply replacing terminology doesn't fix the existential flaw. This distinguishes superficial editing from genuine topological correction in scholarly work.

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