Connectedness and Compactness Questions and Answers

Answer: B

X is connected, but not compact.

X is the union of the unit circles centred at \((n, 0)\), \(n \in \mathbf{Z}\). Adjacent circles, centred 1 apart with radius 1, intersect, so the union of this chain of connected sets is connected.

X is not compact because it is unbounded: it contains the points \((n, 0)\) for every integer \(n\). By the Heine-Borel theorem a compact subset of \(\mathbf{R}^{2}\) must be bounded. So the options calling X compact, or not connected, are wrong.

Answer: A

\(f(A)\) is a connected subset of R³.

A is the closed square \([-1, 1] \times [-1, 1]\), which is connected (indeed convex) and compact. A linear map is continuous, and the continuous image of a connected set is connected. In fact \(f(A)\) is convex, since a linear map sends line segments to line segments.

The other options fail. \(f(A)\) is the continuous image of a compact set, so it is compact and in particular bounded, not unbounded. The image of R² under a linear map into R³ lies in a subspace of dimension at most 2, a closed set with empty interior, so \(f(A)\) can be neither open in R³ nor dense in it.

Answer: B

The compact set is \(B \setminus A\).

A is the strip \(-1 \le x \lt 1\), \(|y| \le 1\) (the condition \(-1 \le |y|\) is automatic), which is not closed because its right edge \(x = 1\) is missing. B is the closed disk of radius \(1/\sqrt{2} \approx 0.707\) centred at (1, 0).

Every point of B has \(|y| \le 0.707\) and \(x \ge 0.29 \gt -1\), so a point of B fails to lie in A only when \(x \ge 1\). Hence \(B \setminus A = B \cap \{x \ge 1\}\), a closed half-disk: closed and bounded, so compact.

The others are not closed. The points \((1 - 1/n, 0)\) lie in \(A \cap B\) but their limit (1, 0) is not in A, so \(A \cap B\) is not closed. The points \((1 - 1/n, 0.9)\) lie in A but outside B (distance about 0.9 from the centre), and their limit (1, 0.9) lies in neither set, so \(A \setminus B\) and \(A \cup B\) are not closed either.

Answer: C

The true statement is Every convex subset of Rⁿ is connected.

A convex set contains the segment between any two of its points, so it is path connected, and path connected sets are connected.

The converse fails, which rules out the first option: a circle, or an annulus, is connected but not convex. Convex sets need not be compact: Rⁿ itself and any open ball are convex but not compact. The union of the closed first and third quadrants is connected (both contain the origin) but not convex: the points (1, 0) and (0, -1) lie in it while their midpoint (1/2, -1/2) lies in the fourth quadrant.

Answer: B

\(\mathbf{R}^{2} - A\) is connected.

A is the integer lattice, a countable closed discrete set. Removing it from the plane leaves a path connected set: any two points can be joined by a polygonal path that dodges the lattice points, for instance by moving along horizontal and vertical lines at non-integer heights and abscissas.

The parabola B behaves differently. \(\mathbf{R}^{2} - B\) splits into the two regions \(y \gt x^{2}\) and \(y \lt x^{2}\), and any path between them must cross the parabola by the intermediate value theorem applied to \(y - x^{2}\). So it is not connected.

Neither complement is closed: A and B are closed and nonempty, so their complements are open proper subsets of the connected plane, and such sets are never closed.

Answer: B

The true statement is A ∪ B is connected if A ∩ B ≠ φ.

If A and B are connected and share a point p, suppose \(A \cup B\) is split into two disjoint open pieces. Each of A and B, being connected, lies entirely in one piece, and both contain p, so both lie in the same piece and the other piece is empty. Hence the union is connected.

The others fail. In the plane, an annulus and a horizontal strip are connected, but their intersection is two separate pieces. A \ B can be disconnected: take A = [0, 2] and B = {1}, with B inside A, which leaves [0, 1) and (1, 2]. Taking B not inside A does not help either: in the plane, a disc minus a line segment that crosses it and runs out beyond it falls into two pieces.

Answer: D

The false statement is A is not connected.

The set \(xy = 0\) is the union of the two coordinate axes. It is the zero set of the continuous function \(xy\), so it is closed in \(\mathbf{R}^{2}\).

Any point on the x-axis can be joined to the origin along that axis, and the origin to any point on the y-axis along the other, so A is path connected and hence connected. That makes 'A is not connected' the false statement.

Answer: C

The disconnected space is S' \(\setminus \{P_{1}, P_{2}\} \times\) [0, 1].

Removing two distinct points from a circle leaves two disjoint open arcs. A product with the connected interval [0, 1] keeps the two pieces apart, so the space is the union of two disjoint non-empty open sets and is not connected.

The others are connected. S' × [0, 1] is a closed annulus (a cylinder), a product of connected spaces. Removing one point from the circle leaves a single open arc, homeomorphic to an interval, so S' \(\setminus\) {P} × [0, 1] is connected. S' × {0} is just a copy of the circle.

Answer: D

X is neither compact nor connected.

\(x^{2} y^{2} = 1\) means \(xy = 1\) or \(xy = -1\), so X is the union of two rectangular hyperbolas, four separate branches in the four quadrants. No branch meets the axes (on the axes \(xy = 0\)), so the branches lie in different open quadrants and X is not connected.

X is closed but unbounded (it contains \((t, 1/t)\) for every \(t \gt 0\)), so by Heine-Borel it is not compact. Each of the other options claims compactness or connectedness, and both fail.

Answer: D

The conclusion is \(|X| \gt 0\): the intersection is non-empty.

Write \(I_{n} = [a_{n}, b_{n}]\). Nesting makes \(a_{n}\) increasing and \(b_{n}\) decreasing, with \(a_{n} \le b_{m}\) for all n, m. So \(a = \sup a_{n}\) exists and satisfies \(a_{n} \le a \le b_{n}\) for every n, which puts a in every \(I_{n}\). This is the nested interval property (Cantor's intersection theorem).

The other options are not forced. X is never empty, so option A fails. X can be a single point (\(I_{n} = [-\frac{1}{n}, \frac{1}{n}]\)) or a whole interval (\(I_{n} = [-1 - \frac{1}{n}, 1 + \frac{1}{n}]\), with intersection [−1, 1]), so neither |X| = 1 nor \(|X| = \infty\) holds in general.