Connectedness and Compactness Questions and Answers

Answer: A

The correct statements are P and R.

P is true: the intervals \((-1/n, 1/n)\) are nested and all contain 0, so every finite subfamily has non-empty intersection. R is true: in a \(T_{1}\) space every singleton is closed, and a finite union of closed sets is closed.

Q is false: with the discrete metric every singleton is an open ball of radius \(1/2\), and these open sets cover the infinite set \(X\) with no finite subcover, so \((X, d)\) is not compact. S is false: a constant map from \(R\) into any Hausdorff space is continuous but not one-one. Any option containing Q or S is therefore wrong.

Answer: D

Both \(f(X)\) and \(X\) are uncountable.

A continuous image of a connected space is connected, and the connected subsets of \(\mathbb{R}\) are intervals. Since \(f\) is not constant, \(f(X)\) contains two different values and hence the whole interval between them, which is uncountable.

\(f\) maps \(X\) onto \(f(X)\), so \(X\) has at least as many points as \(f(X)\) and is uncountable too.

This rules out every option in which \(f(X)\) or \(X\) is countable.

Answer: D

Neither \(S\) nor \(T\) is closed.

As \(x \to 0^{+}\), \(\sin\frac{1}{x}\) takes every value in \([-1, 1]\) infinitely often. So every point \((0, b)\) with \(-1 \le b \le 1\) is a limit point of \(S\), for instance \((0, 1)\) as the limit of the points with \(x = \frac{1}{\pi/2 + 2k\pi}\).

None of these points lies in \(S\), so \(S\) is not closed. \(T\) adds only the origin and still misses \((0, 1)\), so \(T\) is not closed either. Its closure is \(S\) together with the whole segment \(\{0\} \times [-1, 1]\).

Answer: B

The answer is Both P and Q are TRUE.

Both follow from one fact: a continuous bijection from a compact space onto a Hausdorff space is a homeomorphism. \(([0,1], T_{1})\) is compact and Hausdorff.

P: if \(T_{2}\) strictly contains \(T_{1}\) and \(([0,1], T_{2})\) were compact, the identity from \(T_{2}\) to \(T_{1}\) would be a continuous bijection from a compact space to a Hausdorff one, hence a homeomorphism, forcing \(T_{2} = T_{1}\). So \(T_{2}\) is not compact.

Q: if \(T_{2}\) is strictly contained in \(T_{1}\) and were Hausdorff, the identity from \(T_{1}\) to \(T_{2}\) would be a homeomorphism for the same reason, forcing equality. So \(T_{2}\) is not Hausdorff.

A compact Hausdorff topology is both maximal compact and minimal Hausdorff, so every option calling P or Q false is wrong.

Answer: B

The proper map is \(f(x,y) = (x+y, y)\) on \(\mathbb{R} \times \mathbb{R}\).

This f is linear with inverse \((u, v) \mapsto (u - v, v)\), so it is a homeomorphism of the plane. The preimage of a compact set is its image under the continuous inverse, which is compact. So f is proper.

The others fail. The inclusion of \(\mathbb{R} \setminus \{0\}\) into \(\mathbb{R}\) pulls back [-1, 1] to \([-1,0) \cup (0,1]\), not compact. The projection (x, y) → x pulls back the point {0} to the whole y-axis, unbounded. And \(x^{2}-y^{2}\) pulls back {0} to the two lines y = ±x, also unbounded.

Answer: B

X is connected but not path connected.

X is the topologist's sine curve: the graph B of \(\sin(1/x)\) on \((0, 1/\pi]\) together with the segment A on the \(y\)-axis from \(-1\) to 1. B is connected (a continuous image of an interval), and every point of A is a limit of points of B, because \(\sin(1/x)\) takes every value in \([-1, 1]\) on every interval \((0, \epsilon)\). So \(X = A \cup B\) lies between B and its closure and is connected.

No path joins a point of A to a point of B: a path would have to reach \(x = 0\) while its \(y\)-coordinate oscillates between \(-1\) and 1 infinitely often, which contradicts continuity. So X is not path connected, ruling out A, C and D.

Answer: B

The non-compact sets are (ii) and (iii).

In \(\mathbf{R}^{2}\), compact means closed and bounded (Heine-Borel). A parabola is closed but unbounded, so not compact. S, the graph of \(\sin(1/x)\) on \((0, 1]\), is bounded but not closed: its closure adds the segment from \((0, -1)\) to \((0, 1)\), which S does not contain.

A circle is closed and bounded, so compact. T, \(|x| + |y| = 4\), is the boundary of a square (a diamond) through \((\pm 4, 0)\) and \((0, \pm 4)\): closed and bounded, so also compact. That rules out every option containing (i) or (iv).

Answer: C

The true statement is every component of an open set of X is open in X.

This is a standard characterisation: X is locally connected if and only if the components of every open subset are open. If U is open and C is a component of U, each point of C has a connected open neighbourhood inside U, which must lie in C, so C is open.

Local connectedness does not imply connectedness (two disjoint intervals are locally connected but not connected). Nor does it give local path connectedness or make components path components: Kirch's topology on the positive integers is a countable, connected, locally connected Hausdorff space, and a countable Hausdorff space has no non-constant paths, so its single component is not path connected.

Answer: B

The answer is only (b) is true.

(a) is false: [0, 1] is compact and the continuous image of a compact set is compact, but (0, 1) is not compact. (Equivalently, a continuous function on [0, 1] attains its maximum and minimum, so its image is a closed interval.)

(b) is true: for example \(f(x) = \tfrac{1}{2}(1 + \sin(4\pi x))\) is continuous on (0, 1) and takes the values 0 and 1 inside the interval, at \(x = 3/8\) and \(x = 1/8\), so its image is all of [0, 1]. Compactness is not needed in the domain to reach a compact image.

Answer: C

The answer is Y is compact, but X is not compact.

By Heine-Borel, a subset of \(\mathbf{R}^{2}\) is compact exactly when it is closed and bounded. Both sets are closed, being level sets of continuous functions.

Y, defined by \(|x| + |y| = 1\), is the square with vertices \((\pm 1, 0)\) and \((0, \pm 1)\); it is bounded, so it is compact. X, defined by \(|x|\,|y| = 1\), is the four branches of the hyperbolas \(xy = \pm 1\); it contains \((n, 1/n)\) for every \(n\), so it is unbounded and not compact.