Testing a Planar Set for Openness, Compactness, and Connectedness

Question

Let $S = \left\{ (x,y) \in \mathbb{R}^{2} : 1 \le x^{2} + y^{2} < 4 \right\}$ . Which statement about $S$ is true?

a. $S$ is open in $\mathbb{R}^{2}$ .
b. $S$ is compact.
c. $S$ is connected but not compact.
d. $S$ is closed but unbounded.

Correct answer

C. $S$ is connected but not compact.

Full reasoning

1Bounded: every point of $S$ satisfies $x^{2}+y^{2} < 4$ , so $S$ lies inside the disc of radius $2$ and is bounded.
2Not open: the point $(1,0)$ lies in $S$ , but every ball around it contains points with $x^{2}+y^{2} < 1$ , which are not in $S$ .
3Not closed: the sequence $\left(2 - \tfrac{1}{n}, 0\right)$ lies in $S$ for $n \ge 1$ and converges to $(2,0)$ , which is not in $S$ because $2^{2}+0^{2} = 4$ is not less than $4$ . So $S$ omits a limit point.
4Not compact: in $\mathbb{R}^{2}$ , Heine-Borel says compact means closed and bounded, and $S$ fails the closed half.
5Connected: any two points of $S$ can be joined inside $S$ by moving radially to the circle of radius $\tfrac{3}{2}$ (which stays in $S$ , since $1 \le \tfrac{9}{4} < 4$ ) and then along that circle. Path-connected sets are connected, so only the third statement survives.

Why each choice is right or wrong

Choice A

The inner boundary belongs to $S$ : the point $(1,0)$ satisfies $x^{2}+y^{2} = 1$ , and every ball around it contains points with $x^{2}+y^{2} < 1$ , which are outside $S$ . So $(1,0)$ is not an interior point and $S$ is not open.

Choice B

$S$ is bounded, but it is not closed: $(2,0)$ is a limit point of $S$ that $S$ omits, because points just inside the outer circle belong to $S$ and converge to it. Heine-Borel then rules out compactness.

Choice C

Correct. $S$ is path-connected — travel radially to the circle of radius $\tfrac{3}{2}$ , then along that circle — so it is connected, while the missing outer boundary keeps it from being closed and therefore from being compact.

Choice D

Both halves fail. $S$ is bounded, since every point satisfies $x^{2}+y^{2} < 4$ and so lies within distance $2$ of the origin, and $S$ is not closed because it omits the limit point $(2,0)$ .

Related formula

Heine-Borel in Euclidean space

K \subseteq \mathbb{R}^{n} \text{ compact} \iff K \text{ closed and bounded}

Assumptions: The ambient space is R^n with the usual metric; outside R^n only compact implies closed and bounded.

Sources

Authored for this public explainer registry. It has no database question ID and is not copied from a protected or official exam bank.

Mathematical Analysis (Zakon), Section 3.8: Open and Closed Sets. Neighborhoods — LibreTexts Mathematics. Accessed 2026-08-15. Zakon's Mathematical Analysis on LibreTexts is CC BY 3.0; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Mathematical Analysis (Zakon), Section 4.6: Compact Sets — LibreTexts Mathematics. Accessed 2026-08-15. Zakon's Mathematical Analysis on LibreTexts is CC BY 3.0; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Mathematical Analysis (Zakon), Section 4.10: Arcs and Curves. Connected Sets — LibreTexts Mathematics. Accessed 2026-08-15. Zakon's Mathematical Analysis on LibreTexts is CC BY 3.0; attribute the author and platform and avoid verbatim reuse beyond short cited references.

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