Metric Spaces Questions and Answers

Answer: A

The false statement is 'Every closed and bounded subset of a metric space is compact'.

Closed and bounded implies compact in \(\mathbf{R}^{n}\) (Heine-Borel), but not in a general metric space. Take \(\mathbf{R}\) with the discrete metric: the whole space is closed and bounded (every distance is at most 1), yet the cover by singletons has no finite subcover. Likewise the closed unit ball of an infinite-dimensional Hilbert space is closed and bounded but not compact.

The other three are true. A compact subset of a metric space is closed (metric spaces are Hausdorff) and bounded (cover it by balls of radius 1). A closed subset of a compact set is compact. A Cartesian product of compact sets is compact (Tychonoff, or directly for finitely many factors).

Answer: B

A metric space is compact if and only if it is complete and totally bounded.

Total boundedness means that for every \(\epsilon \gt 0\) the space is covered by finitely many \(\epsilon\)-balls. With completeness, it gives sequential compactness: from any sequence one extracts a Cauchy subsequence by repeated pigeonholing into finitely many small balls, and completeness makes it converge.

Boundedness is not enough: the closed unit ball of \(l_{2}\), or \(\mathbf{R}\) with the metric \(\min(1, |x - y|)\), is complete and bounded but not compact. 'Closed' adds nothing to a complete space, and countability is unrelated (\([0, 1]\) is compact and uncountable).

Answer: C

\(d(A, B) \gt 0\) is guaranteed when A is closed and B is compact.

The function \(b \mapsto d(b, A)\) is continuous, and it is positive at each \(b \in B\) because \(b \notin A\) and A is closed. A continuous positive function on a compact set attains a positive minimum, so \(d(A, B) \gt 0\).

The other conditions are not enough. Open sets: A = (0, 1), B = (1, 2) are disjoint with distance 0. Closed sets: A = \(\mathbf{N}\) and \(B = \{n + 1/n : n \ge 2\}\) are closed and disjoint but get arbitrarily close. A singleton: A = {0}, B = (0, 1) gives distance 0.

Answer: B

The true statement is if Y is nowhere dense in X, then \(X \backslash Y\) is dense in X (the paper prints 'nowhere in X').

If Y is nowhere dense, its closure \(\bar{Y}\) has empty interior, so the open set \(X \backslash \bar{Y}\) is dense. Since \(X \backslash Y \supseteq X \backslash \bar{Y}\), \(X \backslash Y\) is dense too.

The others fail with \(X = \mathbf{R}\), \(Y = \mathbf{Q}\): Y is dense, yet \(X \backslash Y\) (the irrationals) is dense, not nowhere dense; \(\mathrm{Int}(\mathbf{Q}) = \phi\); and \(\mathrm{Int}(\mathbf{R} \backslash \mathbf{Q}) = \phi\).

Answer: C

The true statement is [0, 1] is not compact.

In the discrete metric every singleton \(\{x\}\) is the open ball of radius 1/2 about \(x\). The singletons of [0, 1] form an open cover with no finite subcover, because [0, 1] is infinite. So [0, 1] is not compact; in a discrete space only finite sets are compact.

The others are false: every subset is both open and closed, so (0, 1) is closed and {1, ..., 100} is open; and every set is bounded, since no two points are more than 1 apart, so \((0, \infty)\) is bounded.

Answer: B

The answer is (-1, 1) is not complete in \(d\), but there is a complete metric on (-1, 1) inducing the same topology.

With the usual metric, \(x_{n} = 1 - 1/n\) is Cauchy but its limit 1 is not in (-1, 1), so the space is not complete.

However, \(h(x) = \tan(\pi x/2)\) is a homeomorphism of (-1, 1) onto \(\mathbf{R}\). Defining \(d'(x, y) = |h(x) - h(y)|\) gives a metric with the same open sets, and (-1, 1) with \(d'\) is isometric to the complete space \(\mathbf{R}\). So completeness depends on the metric, not only on the topology, which rules out both 'every' and 'no' metric options.

Answer: C

Both (a) and (b) are true.

(a) A non-empty open subset of R contains an open interval \((x - \varepsilon, x + \varepsilon)\), and every non-degenerate interval is uncountable, so the set is uncountable.

(b) The connected subsets of R are exactly the intervals. A connected set containing two points \(a \lt b\) therefore contains the whole interval \([a, b]\), which is uncountable.

Since both statements hold, the options saying only one is true, or both false, are wrong.

Answer: D

The distance is positive if one of Y or Z is compact and the other is closed in X.

Suppose Y is compact, Z closed and disjoint. The function \(y \mapsto d(y, Z)\) is continuous and positive on Y (positive because Z is closed and \(y \notin Z\)), so it attains a positive minimum on compact Y. That minimum is \(d(Y, Z)\).

The weaker conditions fail. Disjoint closed sets can be at distance 0: the hyperbola branch \(xy = 1\) and the x-axis in \(\mathbf{R}^{2}\). Compactness of one set alone is not enough either: \(Y = \{0\}\) and \(Z = (0, 1]\) are disjoint, Y is compact, and \(d(Y, Z) = 0\).

Answer: C

\(f\) maps dense subsets of X to dense subsets of Y.

Let D be dense in X. Continuity gives \(f(\overline{D}) \subseteq \overline{f(D)}\), so \(Y = f(X) = f(\overline{D}) \subseteq \overline{f(D)}\), using surjectivity. Hence \(f(D)\) is dense in Y.

The others fail. \(f(x) = x^{2}\) from R onto \([0, \infty)\) maps the open set \((-1, 1)\) to \([0, 1)\), which is not open (A). The projection \((x, y) \mapsto x\) from \(\mathbf{R}^{2}\) onto R maps the closed hyperbola \(xy = 1\) onto \(\mathbf{R} - \{0\}\), which is not closed (B), and it maps the nowhere dense x-axis onto all of R (D).

Answer: B

The set where two continuous maps agree is closed.

Consider the map \(h : X \to Y \times Y\), \(h(x) = (f(x), g(x))\), which is continuous because both components are. The set \(\{x \mid f(x) = g(x)\}\) is \(h^{-1}(\Delta)\), where \(\Delta = \{(y, y)\}\) is the diagonal of \(Y \times Y\). In a metric space, indeed in any Hausdorff space, the diagonal is closed, so its preimage under a continuous map is closed.

A direct argument works too: if \(x_{n} \to x\) with \(f(x_{n}) = g(x_{n})\), continuity gives \(f(x) = \lim f(x_{n}) = \lim g(x_{n}) = g(x)\).

The set need not be open (take \(f(x) = x\), \(g(x) = 0\) on R: the set is {0}), connected (\(f(x) = x^{2}\), \(g(x) = 1\) gives {-1, 1}) or compact (\(f = g\) gives all of R).