Metric Spaces Questions and Answers

Q: Which of the following statements are TRUE?P : The set \(\{x \in R : |\cos x| \le \frac{1}{2}\}\) is compact.Q : The set \(\{x \in R : \tan x\) is not differentiable\(\}\) is complete.R : The set \(\left\{x \in R : \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n+1}}{(2n+1)!}\right.\) is convergent\(\}\) is bounded.S : The set \(\{x \in R : f(x) = \cos x\) has a local maxima\(\}\) is closed.

The true statements are Q and S.Q: \(\tan x\) fails to be differentiable exactly at \(x = \pi/2 + k\pi\). That set is closed in \(R\) (its points are isolated with no limit point outside it), and a closed subset of the complete space \(R\) is complete. S: \(\cos x\) has its local maxima at \(x = 2k\pi\), again a closed discrete set.P is false: \(|\cos x| \le \frac{1}{2}\) holds on infinitely many intervals \([\pi/3 + k\pi, 2\pi/3 + k\pi]\), so the set is closed but unbounded, and not compact. R is false: the series is the Maclaurin series of \(\sin x\), which converges for every real \(x\), so the set is all of \(R\) and is unbounded. Any option containing P or R is wrong.

Q: Let \(d_{1}\) and \(d_{2}\) denote the usual metric and the discrete metric on \(\mathbb{R}\), respectively. Let \(f : (\mathbb{R}, d_{1}) \to (\mathbb{R}, d_{2})\) be defined by \(f(x) = x,\ x \in \mathbb{R}\). Then

\(f^{-1}\) is continuous but \(f\) is NOT continuous.In the discrete metric every set is open, in particular every singleton \(\{a\}\). Its preimage under \(f\) is the singleton \(\{a\}\) in the usual metric, which is not open. So \(f\) is not continuous.\(f^{-1} : (\mathbb{R}, d_{2}) \to (\mathbb{R}, d_{1})\) is the identity from a discrete space. Every function on a discrete space is continuous, because every preimage is automatically open.That rules out the options saying \(f\) is continuous, and the one saying neither map is continuous.

Q: Let \(X\) and \(Y\) be metric spaces, and let \(f : X \to Y\) be a continuous map. For any subset \(S\) of \(X\), which one of the following statements is true?

The true statement is: if \(S\) is connected, then \(f(S)\) is connected.A continuous image of a connected set is connected: if \(f(S)\) split into two disjoint relatively open pieces, their preimages would split \(S\) the same way.The others fail. Open: \(f(x) = 0\) on \(\mathbb{R}\) sends the open set \(\mathbb{R}\) to \(\{0\}\), which is not open. Closed: \(f(x) = e^{x}\) sends the closed set \(\mathbb{R}\) to \((0, \infty)\), which is not closed. Bounded: \(f(x) = 1/x\) on \(X = (0, 1)\) sends the bounded set \((0, 1)\) to \((1, \infty)\). Continuity preserves compactness and connectedness, not openness, closedness or boundedness.

Q: Consider \(\mathbb{R}^{2}\) with the usual metric. Let \(A = \{(x,y) \in \mathbb{R}^{2} : x^{2} + y^{2} \leq 1\}\) and \(B = \{(x,y) \in \mathbb{R}^{2} : (x-2)^{2} + y^{2} \leq 1\}\). Let \(M = A \cup B\) and \(N = \text{interior}(A) \cup \text{interior}(B)\). Then, which of the following statements is TRUE?

The answer is M is connected and N is not connected.A is the closed unit disc at the origin and B the closed unit disc centred at (2, 0). The centres are 2 apart, equal to the sum of the radii, so the discs touch at exactly one point, (1, 0).M = A ∪ B is a union of two connected sets sharing the point (1, 0), so it is connected. N is the union of the two open discs, which do not contain (1, 0) and are disjoint open sets; they separate N, so N is not connected.That rules out 'M and N are connected' and 'N is connected', and 'neither is connected' fails because M is connected.

Q: Let \((\mathbb{Q}, d)\) be the metric space with d(x, y) = |x-y|. Let \(E = \{p \in \mathbb{Q} : 2 \lt p^{2} \lt 3\}\). Then, the set E is

The set E is closed but not compact.In \(\mathbb{Q}\), E is the set of rationals in \((-\sqrt{3}, -\sqrt{2}) \cup (\sqrt{2}, \sqrt{3})\). The endpoints are irrational, so no rational lies on them and\[ \begin{aligned} E = \mathbb{Q} \cap \big(&[-\sqrt{3}, -\sqrt{2}] \\ &\cup [\sqrt{2}, \sqrt{3}]\big) \end{aligned} \]That is the intersection of \(\mathbb{Q}\) with a closed subset of \(\mathbb{R}\), so E is closed in \(\mathbb{Q}\).E is not compact: a sequence of rationals in E increasing to \(\sqrt{3}\) has no subsequence converging to a point of E, because its only possible limit, \(\sqrt{3}\), is not rational. Compactness would require E to be complete, and it is not. So 'compact' and 'not closed but compact' fail, and 'neither closed nor compact' fails because E is closed.

Q: Consider \(\mathbb{R}^{2}\) with the usual Euclidean metric. Let\[ \begin{aligned} X = &\{(x, x\sin\tfrac{1}{x}) \in \mathbb{R}^{2} : x \in (0,1]\} \\ &\cup \{(0,y) \in \mathbb{R}^{2} : -\infty \lt y \lt \infty\} \end{aligned} \]and\[ \begin{aligned} Y = &\{(x, \sin\tfrac{1}{x}) \in \mathbb{R}^{2} : x \in (0,1]\} \\ &\cup \{(0,y) \in \mathbb{R}^{2} : -\infty \lt y \lt \infty\}. \end{aligned} \]Consider the following statements:P: X is a connected subset of \(\mathbb{R}^{2}\).Q: Y is a connected subset of \(\mathbb{R}^{2}\).Then

Both P and Q are TRUE.Use the fact that if A is connected and \(A \subseteq B \subseteq \overline{A}\), then B is connected, and that two connected sets with a common point have a connected union.X: the graph of \(x\sin(1/x)\) on (0, 1] is connected (a continuous image of an interval) and tends to (0, 0) as \(x \to 0\). Adding the point (0, 0) keeps it connected, and the y-axis, which is connected and contains (0, 0), joins it. So X is connected.Y: the graph of \(\sin(1/x)\) has closure containing the segment \(\{0\} \times [-1, 1]\), so the graph plus that segment is connected (the topologist's sine curve). The whole y-axis is connected and contains the segment, so Y is connected too. Y is not path-connected, but connectedness is all that is asked.

Q: Let \(C[0,1] = \{f : [0,1] \to \mathbb{R} : f \text{ is continuous}\}\).Consider the metric space \((C[0,1], d_{\infty})\), where \(d_{\infty}(f,g) = \sup\{|f(x)-g(x)| : x \in [0,1]\}\) for \(f, g \in C[0,1]\).Let \(f_{0}(x) = 0\) for all \(x \in [0,1]\) and \(X = \{f \in (C[0,1], d_{\infty}) : d_{\infty}(f_{0}, f) \ge \frac{1}{2}\}\).Let \(f_{1}, f_{2} \in C[0,1]\) be defined by \(f_{1}(x) = x\) and \(f_{2}(x) = 1-x\) for all \(x \in [0,1]\).Consider the following statements:P: \(f_{1}\) is in the interior of X.Q: \(f_{2}\) is in the interior of X.Which of the following statements is correct?

Both P and Q are TRUE.X is the set of continuous functions whose supremum norm is at least 1/2. Both \(f_{1}(x) = x\) and \(f_{2}(x) = 1-x\) have norm 1 (at x = 1 and x = 0 respectively).Take any g with \(d_{\infty}(g, f_{1}) \lt \tfrac{1}{2}\). Then \(|g(1)| \gt 1 - \tfrac{1}{2} = \tfrac{1}{2}\), so \(\|g\|_{\infty} \gt \tfrac{1}{2}\) and g is in X. The open ball of radius 1/2 around \(f_{1}\) lies inside X, so \(f_{1}\) is an interior point. The same argument at x = 0 shows \(f_{2}\) is interior.A boundary point of X would need norm exactly 1/2; neither function is near that, so neither is on the boundary.

Q: Consider the metrics \(\rho_{1}\) and \(\rho_{2}\) on \(\mathbb{R}\), defined by\[ \rho_{1}(x,y) = |x-y| \text{ and } \rho_{2}(x,y) = \begin{cases} 0, & \text{if } x = y \\ 1, & \text{if } x \ne y \end{cases} \]Let \(X = \{n \in \mathbb{N} : n \ge 3\}\) and \(Y = \{n + \frac{1}{n} : n \in \mathbb{N}\}\). Define \(f : X \cup Y \to \mathbb{R}\) by\[ f(x) = \begin{cases} 2, & \text{if } x \in X \\ 3, & \text{if } x \in Y \end{cases} \]Consider the following statements:P: The function \(f : (X \cup Y, \rho_{1}) \to (\mathbb{R}, \rho_{1})\) is uniformly continuous.Q: The function \(f : (X \cup Y, \rho_{2}) \to (\mathbb{R}, \rho_{1})\) is uniformly continuous.Then

P is FALSE and Q is TRUE.P: with the usual metric, the points n in X and n + 1/n in Y (for n at least 3) are only 1/n apart, yet f takes the values 2 and 3 there. So no single \(\delta\) makes \(|f(x)-f(y)| \lt \tfrac{1}{2}\) whenever \(|x-y| \lt \delta\): f is not uniformly continuous.Q: with the discrete metric \(\rho_{2}\), taking \(\delta = \tfrac{1}{2}\) means \(\rho_{2}(x,y) \lt \delta\) only when x = y, so \(|f(x)-f(y)| = 0\). Every function on a discrete metric space is uniformly continuous.Note that f is continuous in P's setting too, since every point of X ∪ Y is isolated; it is the uniformity that fails.

Q: Consider the metric spaces \(X = (\mathbb{R}, d_{1})\) and \(Y = ([0,1], d_{2})\) with the metrics defined by \(d_{1}(x,y) = |x - y|,\ x, y \in \mathbb{R}\) and \(d_{2}(x,y) = |x - y|,\ x, y \in [0,1]\), respectively. Then, which one of the following is TRUE?

\([0, \tfrac{1}{4})\) is open in Y but not in X.In \(X = \mathbb{R}\), no open interval around 0 lies inside \([0, \tfrac{1}{4})\), because every such interval contains negative numbers. So the set is not open in X.In Y = [0, 1] with the same metric, the open ball of radius 1/4 about 0 is \(\{y \in [0, 1] : |y| \lt \tfrac{1}{4}\} = [0, \tfrac{1}{4})\), and every other point of the set has a small ball inside it. So the set is open in Y; it equals \((-\tfrac{1}{4}, \tfrac{1}{4}) \cap [0, 1]\), open in the subspace topology. Openness depends on the ambient space, which is the point of the question.

Which of the following statements are TRUE?

P : The set \(\{x \in R : |\cos x| \le \frac{1}{2}\}\) is compact.
Q : The set \(\{x \in R : \tan x\) is not differentiable\(\}\) is complete.
R : The set \(\left\{x \in R : \sum_{n=0}^{\infty} \frac{(-1)^{n} x^{2n+1}}{(2n+1)!}\right.\) is convergent\(\}\) is bounded.
S : The set \(\{x \in R : f(x) = \cos x\) has a local maxima\(\}\) is closed.

A.P and Q
B.R and S
C.Q and S
D.P and S

Exam Year(s)

GATE Mathematics - 2012,

Answer: C

The true statements are Q and S.

Q: \(\tan x\) fails to be differentiable exactly at \(x = \pi/2 + k\pi\). That set is closed in \(R\) (its points are isolated with no limit point outside it), and a closed subset of the complete space \(R\) is complete. S: \(\cos x\) has its local maxima at \(x = 2k\pi\), again a closed discrete set.

P is false: \(|\cos x| \le \frac{1}{2}\) holds on infinitely many intervals \([\pi/3 + k\pi, 2\pi/3 + k\pi]\), so the set is closed but unbounded, and not compact. R is false: the series is the Maclaurin series of \(\sin x\), which converges for every real \(x\), so the set is all of \(R\) and is unbounded. Any option containing P or R is wrong.

Discuss in Forum

Let \(d_{1}\) and \(d_{2}\) denote the usual metric and the discrete metric on \(\mathbb{R}\), respectively. Let \(f : (\mathbb{R}, d_{1}) \to (\mathbb{R}, d_{2})\) be defined by \(f(x) = x,\ x \in \mathbb{R}\). Then

A.\(f\) is continuous but \(f^{-1}\) is NOT continuous
B.\(f^{-1}\) is continuous but \(f\) is NOT continuous
C.both \(f\) and \(f^{-1}\) are continuous
D.neither \(f\) nor \(f^{-1}\) is continuous

Exam Year(s)

GATE Mathematics - 2015,

Answer: B

\(f^{-1}\) is continuous but \(f\) is NOT continuous.

In the discrete metric every set is open, in particular every singleton \(\{a\}\). Its preimage under \(f\) is the singleton \(\{a\}\) in the usual metric, which is not open. So \(f\) is not continuous.

\(f^{-1} : (\mathbb{R}, d_{2}) \to (\mathbb{R}, d_{1})\) is the identity from a discrete space. Every function on a discrete space is continuous, because every preimage is automatically open.

That rules out the options saying \(f\) is continuous, and the one saying neither map is continuous.

Discuss in Forum

Let \(X\) and \(Y\) be metric spaces, and let \(f : X \to Y\) be a continuous map. For any subset \(S\) of \(X\), which one of the following statements is true?

A.If \(S\) is open, then \(f(S)\) is open
B.If \(S\) is connected, then \(f(S)\) is connected
C.If \(S\) is closed, then \(f(S)\) is closed
D.If \(S\) is bounded, then \(f(S)\) is bounded

Exam Year(s)

GATE Mathematics - 2018,

Answer: B

The true statement is: if \(S\) is connected, then \(f(S)\) is connected.

A continuous image of a connected set is connected: if \(f(S)\) split into two disjoint relatively open pieces, their preimages would split \(S\) the same way.

The others fail. Open: \(f(x) = 0\) on \(\mathbb{R}\) sends the open set \(\mathbb{R}\) to \(\{0\}\), which is not open. Closed: \(f(x) = e^{x}\) sends the closed set \(\mathbb{R}\) to \((0, \infty)\), which is not closed. Bounded: \(f(x) = 1/x\) on \(X = (0, 1)\) sends the bounded set \((0, 1)\) to \((1, \infty)\). Continuity preserves compactness and connectedness, not openness, closedness or boundedness.

Discuss in Forum

Consider \(\mathbb{R}^{2}\) with the usual metric. Let \(A = \{(x,y) \in \mathbb{R}^{2} : x^{2} + y^{2} \leq 1\}\) and \(B = \{(x,y) \in \mathbb{R}^{2} : (x-2)^{2} + y^{2} \leq 1\}\). Let \(M = A \cup B\) and \(N = \text{interior}(A) \cup \text{interior}(B)\). Then, which of the following statements is TRUE?

A.M and N are connected
B.Neither M nor N is connected
C.M is connected and N is not connected
D.M is not connected and N is connected

Exam Year(s)

GATE Mathematics - 2022,

Answer: C

The answer is M is connected and N is not connected.

A is the closed unit disc at the origin and B the closed unit disc centred at (2, 0). The centres are 2 apart, equal to the sum of the radii, so the discs touch at exactly one point, (1, 0).

M = A ∪ B is a union of two connected sets sharing the point (1, 0), so it is connected. N is the union of the two open discs, which do not contain (1, 0) and are disjoint open sets; they separate N, so N is not connected.

That rules out 'M and N are connected' and 'N is connected', and 'neither is connected' fails because M is connected.

Discuss in Forum

Let \((\mathbb{Q}, d)\) be the metric space with d(x, y) = |x-y|. Let \(E = \{p \in \mathbb{Q} : 2 \lt p^{2} \lt 3\}\). Then, the set E is

A.closed but not compact
B.not closed but compact
C.compact
D.neither closed nor compact

Exam Year(s)

GATE Mathematics - 2022,

Answer: A

The set E is closed but not compact.

In \(\mathbb{Q}\), E is the set of rationals in \((-\sqrt{3}, -\sqrt{2}) \cup (\sqrt{2}, \sqrt{3})\). The endpoints are irrational, so no rational lies on them and

\[ \begin{aligned} E = \mathbb{Q} \cap \big(&[-\sqrt{3}, -\sqrt{2}] \\ &\cup [\sqrt{2}, \sqrt{3}]\big) \end{aligned} \]

That is the intersection of \(\mathbb{Q}\) with a closed subset of \(\mathbb{R}\), so E is closed in \(\mathbb{Q}\).

E is not compact: a sequence of rationals in E increasing to \(\sqrt{3}\) has no subsequence converging to a point of E, because its only possible limit, \(\sqrt{3}\), is not rational. Compactness would require E to be complete, and it is not. So 'compact' and 'not closed but compact' fail, and 'neither closed nor compact' fails because E is closed.

Discuss in Forum

Consider \(\mathbb{R}^{2}\) with the usual Euclidean metric. Let

\[ \begin{aligned} X = &\{(x, x\sin\tfrac{1}{x}) \in \mathbb{R}^{2} : x \in (0,1]\} \\ &\cup \{(0,y) \in \mathbb{R}^{2} : -\infty \lt y \lt \infty\} \end{aligned} \]

and

\[ \begin{aligned} Y = &\{(x, \sin\tfrac{1}{x}) \in \mathbb{R}^{2} : x \in (0,1]\} \\ &\cup \{(0,y) \in \mathbb{R}^{2} : -\infty \lt y \lt \infty\}. \end{aligned} \]

Consider the following statements:

P: X is a connected subset of \(\mathbb{R}^{2}\).
Q: Y is a connected subset of \(\mathbb{R}^{2}\).

Then

A.both P and Q are TRUE
B.P is FALSE and Q is TRUE
C.P is TRUE and Q is FALSE
D.both P and Q are FALSE

Exam Year(s)

GATE Mathematics - 2023,

Answer: A

Both P and Q are TRUE.

Use the fact that if A is connected and \(A \subseteq B \subseteq \overline{A}\), then B is connected, and that two connected sets with a common point have a connected union.

X: the graph of \(x\sin(1/x)\) on (0, 1] is connected (a continuous image of an interval) and tends to (0, 0) as \(x \to 0\). Adding the point (0, 0) keeps it connected, and the y-axis, which is connected and contains (0, 0), joins it. So X is connected.

Y: the graph of \(\sin(1/x)\) has closure containing the segment \(\{0\} \times [-1, 1]\), so the graph plus that segment is connected (the topologist's sine curve). The whole y-axis is connected and contains the segment, so Y is connected too. Y is not path-connected, but connectedness is all that is asked.

Discuss in Forum

Let \(C[0,1] = \{f : [0,1] \to \mathbb{R} : f \text{ is continuous}\}\).

Consider the metric space \((C[0,1], d_{\infty})\), where \(d_{\infty}(f,g) = \sup\{|f(x)-g(x)| : x \in [0,1]\}\) for \(f, g \in C[0,1]\).

Let \(f_{0}(x) = 0\) for all \(x \in [0,1]\) and \(X = \{f \in (C[0,1], d_{\infty}) : d_{\infty}(f_{0}, f) \ge \frac{1}{2}\}\).

Let \(f_{1}, f_{2} \in C[0,1]\) be defined by \(f_{1}(x) = x\) and \(f_{2}(x) = 1-x\) for all \(x \in [0,1]\).

Consider the following statements:

P: \(f_{1}\) is in the interior of X.
Q: \(f_{2}\) is in the interior of X.

Which of the following statements is correct?

A.P is TRUE and Q is FALSE
B.P is FALSE and Q is TRUE
C.Both P and Q are FALSE
D.Both P and Q are TRUE

Exam Year(s)

GATE Mathematics - 2023,

Answer: D

Both P and Q are TRUE.

X is the set of continuous functions whose supremum norm is at least 1/2. Both \(f_{1}(x) = x\) and \(f_{2}(x) = 1-x\) have norm 1 (at x = 1 and x = 0 respectively).

Take any g with \(d_{\infty}(g, f_{1}) \lt \tfrac{1}{2}\). Then \(|g(1)| \gt 1 - \tfrac{1}{2} = \tfrac{1}{2}\), so \(\|g\|_{\infty} \gt \tfrac{1}{2}\) and g is in X. The open ball of radius 1/2 around \(f_{1}\) lies inside X, so \(f_{1}\) is an interior point. The same argument at x = 0 shows \(f_{2}\) is interior.

A boundary point of X would need norm exactly 1/2; neither function is near that, so neither is on the boundary.

Discuss in Forum

Consider the metrics \(\rho_{1}\) and \(\rho_{2}\) on \(\mathbb{R}\), defined by

\[ \rho_{1}(x,y) = |x-y| \text{ and } \rho_{2}(x,y) = \begin{cases} 0, & \text{if } x = y \\ 1, & \text{if } x \ne y \end{cases} \]

Let \(X = \{n \in \mathbb{N} : n \ge 3\}\) and \(Y = \{n + \frac{1}{n} : n \in \mathbb{N}\}\). Define \(f : X \cup Y \to \mathbb{R}\) by

\[ f(x) = \begin{cases} 2, & \text{if } x \in X \\ 3, & \text{if } x \in Y \end{cases} \]

Consider the following statements:

P: The function \(f : (X \cup Y, \rho_{1}) \to (\mathbb{R}, \rho_{1})\) is uniformly continuous.
Q: The function \(f : (X \cup Y, \rho_{2}) \to (\mathbb{R}, \rho_{1})\) is uniformly continuous.

Then

A.P is TRUE and Q is FALSE
B.P is FALSE and Q is TRUE
C.both P and Q are FALSE
D.both P and Q are TRUE

Exam Year(s)

GATE Mathematics - 2023,

Answer: B

P is FALSE and Q is TRUE.

P: with the usual metric, the points n in X and n + 1/n in Y (for n at least 3) are only 1/n apart, yet f takes the values 2 and 3 there. So no single \(\delta\) makes \(|f(x)-f(y)| \lt \tfrac{1}{2}\) whenever \(|x-y| \lt \delta\): f is not uniformly continuous.

Q: with the discrete metric \(\rho_{2}\), taking \(\delta = \tfrac{1}{2}\) means \(\rho_{2}(x,y) \lt \delta\) only when x = y, so \(|f(x)-f(y)| = 0\). Every function on a discrete metric space is uniformly continuous.

Note that f is continuous in P's setting too, since every point of X ∪ Y is isolated; it is the uniformity that fails.

Discuss in Forum

Consider the metric spaces \(X = (\mathbb{R}, d_{1})\) and \(Y = ([0,1], d_{2})\) with the metrics defined by \(d_{1}(x,y) = |x - y|,\ x, y \in \mathbb{R}\) and \(d_{2}(x,y) = |x - y|,\ x, y \in [0,1]\), respectively. Then, which one of the following is TRUE?

A.\(\left[0, \frac{1}{4}\right)\) is open in X but not in Y
B.\(\left[0, \frac{1}{4}\right)\) is open in Y but not in X
C.\(\left[0, \frac{1}{4}\right)\) is open in both X and Y
D.\(\left[0, \frac{1}{4}\right)\) is open neither in X nor in Y

Exam Year(s)

GATE Mathematics - 2025,

Answer: B

\([0, \tfrac{1}{4})\) is open in Y but not in X.

In \(X = \mathbb{R}\), no open interval around 0 lies inside \([0, \tfrac{1}{4})\), because every such interval contains negative numbers. So the set is not open in X.

In Y = [0, 1] with the same metric, the open ball of radius 1/4 about 0 is \(\{y \in [0, 1] : |y| \lt \tfrac{1}{4}\} = [0, \tfrac{1}{4})\), and every other point of the set has a small ball inside it. So the set is open in Y; it equals \((-\tfrac{1}{4}, \tfrac{1}{4}) \cap [0, 1]\), open in the subspace topology. Openness depends on the ambient space, which is the point of the question.

Discuss in Forum

10.

Let X be a set with at least two elements. Let \(d_1\), \(d_2\) and \(d_3\) be metrics on X. Which one of the following is NOT a metric on X?

A.\(d(x, y) := \min(d_1(x, y), 3)\) for all \(x, y \in X\).
B.\(d(x, y) := \max(d_2(x, y), 3)\) for all \(x, y \in X\).
C.\(d(x, y) := \frac{10\, d_3(x,y)}{1 + d_3(x,y)}\) for all \(x, y \in X\).
D.\(d(x, y) := \frac{1}{3}(d_1(x, y) + d_2(x, y) + d_3(x, y))\) for all \(x, y \in X\).

Exam Year(s)

GATE Mathematics - 2026,

Answer: B

The one that is not a metric is \(d(x, y) := \max(d_2(x, y), 3)\).

A metric must give \(d(x, x) = 0\). Here \(d(x, x) = \max(0, 3) = 3\), so it fails at once.

The others are metrics. Truncating a metric with \(\min(d_1, 3)\) keeps the triangle inequality. \(\frac{10 d_3}{1 + d_3}\) is a positive multiple of the standard bounded metric \(\frac{d}{1+d}\), built from an increasing, subadditive function. An average of metrics, \(\frac{1}{3}(d_1 + d_2 + d_3)\), is again a metric, since each property survives addition and scaling.

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