Metric Spaces Questions and Answers

Answer: B

The union \(A \cup B\) is dense.

A set is dense when its closure is the whole space. Since \(A \subseteq A \cup B\), the closure of \(A \cup B\) contains the closure of A, which is already the whole space. So the union of a dense set with anything is dense.

The other options fail. In R, the rationals A and the irrationals B are both proper dense subsets, but \(A \cap B = \phi\), which is not dense, and \(A \setminus B = \mathbf{Q}\) is dense here but fails if instead \(B = \mathbf{R} \setminus \{0\}\) and \(A = \mathbf{R} \setminus \{1\}\), where \(A \setminus B = \{0\}\). Neither set need be closed: a closed dense set is the whole space, and both A and B are proper.

Answer: D

Neither \(d_{1}\) nor \(d_{2}\) is a metric on R.

A metric must vanish only on equal points. Both functions see only \(|x|\):

\[ \begin{aligned} d_{1}(1, -1) &= \big|\,|1| - |-1|\,\big| = 0 \\ d_{2}(1, -1) &= |1 - 1| = 0 \end{aligned} \]

yet \(1 \neq -1\). So both fail the identity of indiscernibles. (Each is symmetric and satisfies the triangle inequality, so each is a pseudometric; they become metrics only on \([0, \infty)\).)

That rules out every option claiming that one or both are metrics. The trap is that \(d_{1}\) looks like the familiar \(|x - y|\) and passes the triangle inequality check.

Answer: B

K must be compact.

In \(\mathbf{R}^{n}\) a set is compact exactly when it is closed and bounded. If K were not bounded, \(f(x) = \|x\|\) would be a continuous unbounded function on K. If K were bounded but not closed, there would be a limit point \(a\) of K outside K, and \(f(x) = 1/\|x - a\|\) would be continuous and unbounded on K. So the property forces K to be closed and bounded, that is, compact; conversely every continuous function on a compact set is bounded.

'Bounded' alone is not enough (the open interval (0, 1) carries the unbounded \(1/x\)), and connectedness or density has nothing to do with it.

Answer: C

The true statement is \(|X| \neq 10\).

In a metric space every finite set is closed, and every single point of a finite space is also open (its complement is a finite union of closed points). So a finite metric space with two or more points splits into open, closed singletons and is disconnected. A connected metric space is therefore either a single point or infinite; in fact a connected metric space with two points \(a \neq b\) contains a point at every distance \(r \lt d(a, b)\) from \(a\), so it is uncountable.

Either way it cannot have exactly 10 points. 'X is finite' fails for the real line, 'X is infinite' and 'X is uncountable' fail for a one-point space, so only \(|X| \neq 10\) holds for every connected metric space.

Answer: A

The functions agree everywhere if D is dense in X.

The set \(D = \{x : f(x) = g(x)\}\) is always closed: if \(x_{n} \in D\) and \(x_{n} \to x\), continuity gives \(f(x) = \lim f(x_{n}) = \lim g(x_{n}) = g(x)\). A closed set that is dense is the whole space, so \(D = X\) and \(f \equiv g\).

The other properties are not enough. On \(X = \mathbf{R}\), \(f(x) = x\) and \(g(x) = 0\) agree only on \(D = \{0\}\), which is compact and closed; and \(f(x) = \sin x\), \(g(x) = 0\) agree on the multiples of \(\pi\), an unbounded set, while still being different functions.