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Exercise Set 1 (Homework 01 & 02)

This section contains the problems and detailed solutions for the first and second homework assignments of the Point-Set Topology course.

Problem 1

Let \(A\) be a non-empty finite set. Let \(\mathcal{P}(A) = \{A' \mid A' \subseteq A\}\) be the power set of \(A\). Let \(2^A\) denote the set of all functions of the form \(f: A \to \{\pm 1\}\).

Prove: There exists a bijection between \(\mathcal{P}(A)\) and \(2^A\), and calculate the cardinality of \(\mathcal{P}(A)\).

Solution

We can construct a bijection by defining an indicator function (characteristic function).

1. Construct the mapping: Define the mapping \(\Phi: \mathcal{P}(A) \to 2^A\). For any subset \(A' \in \mathcal{P}(A)\), its corresponding function \(f_{A'} = \Phi(A')\) is defined as follows:

\[ f_{A'}(x) = \begin{cases} 1, & x \in A' \\ -1, & x \notin A' \end{cases} \]

2. Prove \(\Phi\) is injective: Assume that for \(A_1', A_2' \in \mathcal{P}(A)\), we have \(\Phi(A_1') = \Phi(A_2')\). This means that for all \(x \in A\), \(f_{A_1'}(x) = f_{A_2'}(x)\). According to the definition of \(f_{A'}\): If \(x \in A_1'\), then \(f_{A_1'}(x) = 1\), and thus \(f_{A_2'}(x) = 1\), which by definition implies \(x \in A_2'\). Therefore, \(A_1' \subseteq A_2'\). Similarly, we can obtain \(A_2' \subseteq A_1'\). Therefore \(A_1' = A_2'\), so the mapping is injective.

3. Prove \(\Phi\) is surjective: For any function \(g \in 2^A\), where \(g: A \to \{\pm 1\}\). We can construct the set \(A_g' = \{x \in A \mid g(x) = 1\}\). Clearly \(A_g' \subseteq A\), i.e., \(A_g' \in \mathcal{P}(A)\). By construction, the function generated by \(\Phi(A_g')\) takes \(1\) when \(x \in A_g'\), and \(-1\) otherwise, which is exactly consistent with the values of \(g\). Therefore \(\Phi(A_g') = g\), so the mapping is surjective.

4. Calculate the cardinality: Since a bijection exists, \(|\mathcal{P}(A)| = |2^A|\). For a finite set \(A\), let \(|A| = n\). When constructing the function \(f: A \to \{-1, 1\}\), for every element \(x\) in \(A\), there are \(2\) value choices (\(1\) or \(-1\)). According to the multiplication principle, the total number of functions is:

\[ |2^A| = \underbrace{2 \times 2 \times \dots \times 2}_{n \text{ times}} = 2^n \]

Therefore, the cardinality of the power set is \(|\mathcal{P}(A)| = 2^{|A|}\).

Problem 2

Let \(A\) be a non-empty finite set. Let \(\mathcal{F}(A)\) denote the set of all bijections \(f: A \to A\).

Prove: \(\mathcal{F}(A)\) forms a group under the operation of mapping composition \(g \circ f\). Also discuss under what conditions \(\mathcal{F}(A)\) is an abelian group.

Solution

To prove that \((\mathcal{F}(A), \circ)\) forms a group, we need to verify the four group axioms:

1. Closure: If \(f, g \in \mathcal{F}(A)\), then \(f\) and \(g\) are both bijections. Since the composition of two bijections remains a bijection, \(g \circ f: A \to A\) belongs to \(\mathcal{F}(A)\).

2. Associativity: For any \(f, g, h \in \mathcal{F}(A)\), according to the properties of mapping composition, for any \(x \in A\):

\[ [h \circ (g \circ f)](x) = h((g \circ f)(x)) = h(g(f(x))) \]
\[ [(h \circ g) \circ f](x) = (h \circ g)(f(x)) = h(g(f(x))) \]

Therefore \(h \circ (g \circ f) = (h \circ g) \circ f\).

3. Identity Element: There exists an identity mapping \(id_A: A \to A\), defined as \(id_A(x) = x\). Clearly, \(id_A\) is a bijection and belongs to \(\mathcal{F}(A)\). For any \(f \in \mathcal{F}(A)\), we have \(f \circ id_A = id_A \circ f = f\).

4. Inverse Element: For any \(f \in \mathcal{F}(A)\), since \(f\) is a bijection, according to the inverse mapping theorem, there exists a unique inverse mapping \(f^{-1}: A \to A\) and \(f^{-1}\) is also a bijection. Therefore \(f^{-1} \in \mathcal{F}(A)\), and it satisfies \(f \circ f^{-1} = f^{-1} \circ f = id_A\).

5. Discussion on Abelian Group: \(\mathcal{F}(A)\) is an abelian group if and only if \(|A| \le 2\).

  • If \(|A| = 1\), \(\mathcal{F}(A)\) contains only the identity element and is an abelian group.

  • If \(|A| = 2\), let \(A = \{a, b\}\), \(\mathcal{F}(A)\) contains \(id_A\) and the transposition \((a, b)\); its order is \(2\), and the commutative law holds.

  • If \(|A| \ge 3\), \(\mathcal{F}(A)\) is the symmetric group \(S_n\) (\(n \ge 3\)). In this case, we can take the transpositions \(f = (1, 2)\) and \(g = (2, 3)\). Then \((g \circ f)(1) = g(2) = 3\), whereas \((f \circ g)(1) = f(1) = 2\). The two are not equal, so it is not an abelian group.

Problem 3

Let \(p_n\) be the \(n\)-th prime number arranged in ascending order (e.g., \(p_1=2, p_2=3, p_3=5\)). Prove by mathematical induction or other methods: for every \(n\), \(p_n < 2^{2^n}\).

Solution

We will use mathematical induction based on the estimation properties of the number of prime numbers.

1. Base Case: When \(n = 1\), \(p_1 = 2\). The right side is \(2^{2^1} = 2^2 = 4\). Clearly \(2 < 4\), so the proposition holds for \(n=1\).

2. Inductive Hypothesis: Assume the proposition holds for \(n \le k\), i.e., for all \(1 \le i \le k\), we have \(p_i < 2^{2^i}\).

3. Inductive Step: Based on Euclid's proof idea for the infinitude of primes, the \((k+1)\)-th prime \(p_{k+1}\) satisfies the following inequality:

\[ p_{k+1} \le (p_1 p_2 \dots p_k) + 1 \]

Substituting the inductive hypothesis:

\[ p_{k+1} \le (2^{2^1} \cdot 2^{2^2} \cdot \dots \cdot 2^{2^k}) + 1 \]

According to the rules of exponents, the exponent part inside the parentheses is the sum of a geometric progression:

\[ 2^1 + 2^2 + \dots + 2^k = 2(2^k - 1) = 2^{k+1} - 2 \]

Therefore:

\[ p_{k+1} \le 2^{2^{k+1} - 2} + 1 \]

Since for \(k \ge 1\), we have \(2^{k+1} - 2 < 2^{k+1}\). And when \(x \ge 2\), \(2^{x-2} + 1 < 2^x\) clearly holds. Specifically, for \(k \ge 1\):

\[ 2^{2^{k+1} - 2} + 1 \le 2^{2^{k+1} - 1} < 2^{2^{k+1}} \]

Hence \(p_{k+1} < 2^{2^{k+1}}\). In summary, by mathematical induction, the proposition holds for all \(n \in \mathbb{N}_{\ge 1}\).

Problem 4

Let \(S\) and \(S'\) be the following subsets of \(\mathbb{R} \times \mathbb{R}\): \(S = \{(x, y) \mid y = x+1 \text{ and } 0 < x < 3\}\) \(S' = \{(x, y) \mid y - x \in \mathbb{Z}\}\)

  • (i) Prove that \(S'\) is an equivalence relation on \(\mathbb{R}\). Describe the equivalence classes of \(S'\).

  • (ii) Let \(\{R_i\}_{i \in I}\) be a family of equivalence relations on \(\mathbb{R}\). Prove that \(\bigcap_{i \in I} R_i\) is also an equivalence relation on \(\mathbb{R}\).

  • (iii) Describe the equivalence relation \(T\) on \(\mathbb{R}\), which is the intersection of all equivalence relations containing \(S\). Describe the equivalence classes of \(T\).

Solution

Solution (i): Prove \(S'\) is an equivalence relation

  • Reflexivity: For any \(x \in \mathbb{R}\), \(x - x = 0\). Since \(0 \in \mathbb{Z}\), we have \((x, x) \in S'\).

  • Symmetry: If \((x, y) \in S'\), then \(y - x = k \in \mathbb{Z}\). Then \(x - y = -k\). Since \(-k\) is also an integer, \((y, x) \in S'\).

  • Transitivity: If \((x, y) \in S'\) and \((y, z) \in S'\), then \(y - x = k_1 \in \mathbb{Z}\) and \(z - y = k_2 \in \mathbb{Z}\). Then \(z - x = (z - y) + (y - x) = k_2 + k_1\). Since the sum of integers is still an integer, \((x, z) \in S'\).

  • Description of equivalence classes: For any \(x \in \mathbb{R}\), its equivalence class is \([x]_{S'} = \{x + k \mid k \in \mathbb{Z}\}\). This essentially discretizes the real number line by integer spans.

Solution (ii): Prove the intersection of equivalence relations is still an equivalence relation Let \(R = \bigcap_{i \in I} R_i\).

  • Reflexivity: For any \(x \in \mathbb{R}\), since each \(R_i\) is an equivalence relation, we have \(\forall i \in I, (x, x) \in R_i\). Therefore \((x, x) \in R\).

  • Symmetry: If \((x, y) \in R\), then for all \(i \in I\), \((x, y) \in R_i\). By the symmetry of \(R_i\), we know \((y, x) \in R_i\). Therefore \((y, x) \in R\).

  • Transitivity: If \((x, y) \in R\) and \((y, z) \in R\), then for all \(i \in I\), \((x, y) \in R_i\) and \((y, z) \in R_i\). By the transitivity of \(R_i\), we know \((x, z) \in R_i\). Therefore \((x, z) \in R\).

Solution (iii): Describe the equivalence relation \(T\) According to the conclusion of (ii), \(T\) is the smallest equivalence relation containing \(S\). The equivalence relation \(T\) is generated by the reflexive, symmetric, and transitive closure of \(S\). \(S\) describes a connection relation with an interval of \(1\) where \(x \in (0, 3)\).

  • Description of equivalence classes:

  • If \(x \in (0, 1]\), it can establish an equivalence relation with subsequent points by continuously adding \(1\); its equivalence class is \([x]_T = \{x, x+1, x+2, x+3\}\) (Note: when \(x\) approaches \(1\), \(x+3\) approaches \(4\)).

  • If \(x \in (1, 4)\) and does not belong to the points in the above generation method, its equivalence class overlaps with the above (i.e., it belongs to a sequence whose starting point is in \((0,1]\)).

  • If \(x \in (-\infty, 0] \cup [4, +\infty)\), it is not "connected" to any other points by the relation \(S\), therefore \([x]_T = \{x\}\).

Problem 5

A real number \(x \in \mathbb{R}\) is called an algebraic number if it satisfies a polynomial equation of positive degree with rational coefficients:

\[ x^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0 = 0 \]

where \(a_i \in \mathbb{Q}\); otherwise, it is called a transcendental number. Assume that each such polynomial equation has only a finite number of roots in \(\mathbb{R}\).

Prove: The set of all real algebraic numbers \(\mathcal{A} \subseteq \mathbb{R}\) is countable. And from this, deduce that the set of all real transcendental numbers \(\mathbb{R} - \mathcal{A}\) is uncountable.

Solution

Step 1: Prove the set of polynomials with rational coefficients is countable For each specific degree \(n \ge 1\), a monic polynomial with rational coefficients is uniquely determined by its coefficient sequence \((a_{n-1}, a_{n-2}, \dots, a_0) \in \mathbb{Q}^n\). Because the set of rational numbers \(\mathbb{Q}\) is countable, the Cartesian product of a finite number of countable sets \(\mathbb{Q}^n\) is also countable. The set of all polynomials of positive degree with rational coefficients can be expressed as the union of sets of polynomials of fixed degrees:

\[ \mathcal{P} = \bigcup_{n=1}^{\infty} \mathbb{Q}^n \]

Since the union of countably many countable sets remains countable, the set \(\mathcal{P}\) of all polynomials with rational coefficients is countable.

Step 2: Prove the set of algebraic numbers \(\mathcal{A}\) is countable For each polynomial \(P \in \mathcal{P}\), let \(R(P)\) be the set of all roots of the polynomial \(P\) in \(\mathbb{R}\). According to the problem's assumption, \(R(P)\) is a finite set. The set of all real algebraic numbers \(\mathcal{A}\) can be expressed as the union of the root sets of all polynomials with rational coefficients:

\[ \mathcal{A} = \bigcup_{P \in \mathcal{P}} R(P) \]

Since \(\mathcal{P}\) is countable and each \(R(P)\) is a finite set, according to the theorem that "the union of countably many countable (or finite) sets is still a countable set," \(\mathcal{A}\) is countable.

Step 3: Deduce the uncountability of the set of transcendental numbers We already know that the set of real numbers \(\mathbb{R}\) is uncountable. Since \(\mathbb{R} = \mathcal{A} \cup (\mathbb{R} - \mathcal{A})\), if the set of transcendental numbers \(\mathbb{R} - \mathcal{A}\) were countable, then the union of two countable sets, \(\mathbb{R}\), should also be countable. This contradicts the uncountability of the set of real numbers. Therefore, the set of real transcendental numbers \(\mathbb{R} - \mathcal{A}\) must be uncountable.

Problem 6

Let \(X := \{0, 1\}\). Let \(X^{\mathbb{N}}\) be the set of all mappings \(f: \mathbb{N} \to X\). For \(f \in X^{\mathbb{N}}\), define the support of \(f\) as:

\[ supp(f) = f^{-1}(\{1\}) \]

Consider the subset \(X_0^{\mathbb{N}} = \{f \in X^{\mathbb{N}} \mid supp(f) \text{ is a finite set} \}\).

Question: Is \(X_0^{\mathbb{N}}\) countable or uncountable? Please provide your reasoning.

Solution

Conclusion: \(X_0^{\mathbb{N}}\) is a countable set.

Reasoning as follows: Every function \(f \in X_0^{\mathbb{N}}\) is uniquely determined by its support set \(supp(f)\). This is because \(f(n)=1\) if and only if \(n \in supp(f)\), otherwise \(f(n)=0\). Therefore, there exists a natural bijection between \(X_0^{\mathbb{N}}\) and the set \(\mathcal{P}_{fin}(\mathbb{N})\) of all finite subsets of the natural numbers \(\mathbb{N}\).

We can partition \(\mathcal{P}_{fin}(\mathbb{N})\) into a family of subsets categorized by size:

\[ \mathcal{P}_{fin}(\mathbb{N}) = \bigcup_{n=0}^{\infty} \mathcal{P}_n(\mathbb{N}) \]

where \(\mathcal{P}_n(\mathbb{N})\) represents the set consisting of all subsets of natural numbers of exact size \(n\).

  • For \(n=0\), \(\mathcal{P}_0(\mathbb{N}) = \{\emptyset\}\), which is a finite set.

  • For \(n \ge 1\), every subset containing \(n\) elements can be mapped injectively to an ordered \(n\)-tuple \((k_1, k_2, \dots, k_n) \in \mathbb{N}^n\) (for example, arranged in ascending order). Since \(\mathbb{N}^n\) is a countable set, its subset \(\mathcal{P}_n(\mathbb{N})\) must also be countable.

Because \(\mathcal{P}_{fin}(\mathbb{N})\) is the union of countably many countable sets, it is countable. Consequently, \(X_0^{\mathbb{N}}\) is a countable set.

Definition 0.1 (Pseudo-metric space)

A pseudo-metric space is a pair \((S, d)\), where \(S\) is a non-empty set, and \(d: S \times S \to \mathbb{R}_{\ge 0}\) is a function (called a pseudo-metric on \(S\)) satisfying the following properties: (1) \(d(x, x) = 0\) (2) For all \(x, y \in S\), \(d(x, y) = d(y, x)\) (3) For all \(x, y, z \in S\), the triangle inequality holds: \(d(x, z) \le d(x, y) + d(y, z)\)

Note: If it additionally satisfies the condition "if \(d(x, y) = 0\) then it must be \(x = y\)", then it becomes a strict metric space.

(The following problems 7, 8, and 9 are based on the related concepts of metrics and pseudo-metrics)

Problem 7

Let \((S, d)\) be a pseudo-metric space. We define a relation \(\sim\) on \(S\) as follows: if \(d(x, y) = 0\), then we say \(x \sim y\).

Prove: (i) \(\sim\) is an equivalence relation on \(S\) (let the quotient set formed by its equivalence classes be denoted as \(S^*\)); (ii) For every \([x], [y] \in S^*\), defining the function \(d^*([x], [y]) = d(x, y)\), prove that it gives a well-defined metric on \(S^*\).

Solution

(i) Prove it is an equivalence relation

  • Reflexivity: From pseudo-metric definition (1), \(d(x, x) = 0\), so \(x \sim x\).

  • Symmetry: If \(x \sim y\), then \(d(x, y) = 0\). From pseudo-metric definition (2), \(d(y, x) = d(x, y) = 0\), so \(y \sim x\).

  • Transitivity: If \(x \sim y\) and \(y \sim z\), then \(d(x, y) = 0\) and \(d(y, z) = 0\). By the triangle inequality: \(0 \le d(x, z) \le d(x, y) + d(y, z) = 0 + 0 = 0\)

Thus \(d(x, z) = 0\), which means \(x \sim z\).

(ii) Prove \(d^*\) is a well-defined metric

  • 1. Well-definedness: We need to prove that the value of \(d^*([x], [y])\) does not depend on the choice of the representative elements of the equivalence classes. Let \(x \sim x'\) and \(y \sim y'\), i.e., \(d(x, x') = 0\) and \(d(y, y') = 0\). According to the triangle inequality: \(d(x, y) \le d(x, x') + d(x', y') + d(y', y) = 0 + d(x', y') + 0 = d(x', y')\)

Similarly, applying the triangle inequality in reverse yields \(d(x', y') \le d(x, y)\). Therefore, \(d(x, y) = d(x', y')\), the value is independent of the representatives, hence well-defined.

  • 2. Verify metric axioms:

  • Non-negativity and Reflexivity: \(d^*([x], [y]) = d(x, y) \ge 0\). And \(d^*([x], [x]) = d(x, x) = 0\).

  • Strict Positivity: If \(d^*([x], [y]) = 0\), then \(d(x, y) = 0\). According to the definition of \(\sim\), this means \(x \sim y\), i.e., \([x] = [y]\).

  • Symmetry: \(d^*([x], [y]) = d(x, y) = d(y, x) = d^*([y], [x])\).

  • Triangle Inequality:

    \[ d^*([x], [z]) = d(x, z) \le d(x, y) + d(y, z) = d^*([x], [y]) + d^*([y], [z]) \]

Thus \(d^*\) is indeed a metric on \(S^*\).

Problem 8

Let \(n \in \mathbb{N}_{\ge 1}\). Consider the function \(d_n: \mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}_{\ge 0}\) defined on \(\mathbb{R}^2 \times \mathbb{R}^2\), whose formula is:

\[ d_n(x, y) := (|x_1 - y_1| + |x_2 - y_2|)^n \]

where \(x = (x_1, x_2), y = (y_1, y_2) \in \mathbb{R}^2\). Determine all values of \(n\) that make \((\mathbb{R}^2, d_n)\) a metric space.

Solution

Conclusion: The only value of \(n\) that makes \((\mathbb{R}^2, d_n)\) a metric space is \(1\).

Analysis and Proof: For \(d_n\) to be a metric, it must satisfy the triangle inequality for all points: \(d_n(x, z) \le d_n(x, y) + d_n(y, z)\).

Case 1: When \(n = 1\)

\[ d_1(x, y) = |x_1 - y_1| + |x_2 - y_2| \]

This is exactly the Manhattan metric (\(L_1\) metric) on \(\mathbb{R}^2\). According to the triangle inequality for absolute values of real numbers:

\[ |x_1 - z_1| + |x_2 - z_2| \le (|x_1 - y_1| + |y_1 - z_1|) + (|x_2 - y_2| + |y_2 - z_2|) \]

Rearranging this yields \(d_1(x, z) \le d_1(x, y) + d_1(y, z)\). Non-negativity, strict positivity, and symmetry clearly hold. Therefore, \(n=1\) is a solution.

Case 2: When \(n \ge 2\) We try to find a counterexample that violates the triangle inequality. Take three points in \(\mathbb{R}^2\): \(x = (0, 0)\), \(y = (1, 0)\), \(z = (1, 1)\).

We calculate the distances between these three points pairwise:

\[ d_n(x, y) = (|0 - 1| + |0 - 0|)^n = 1^n = 1 \]
\[ d_n(y, z) = (|1 - 1| + |0 - 1|)^n = 1^n = 1 \]
\[ d_n(x, z) = (|0 - 1| + |0 - 1|)^n = (1 + 1)^n = 2^n \]

To satisfy the triangle inequality, we must have:

\[ d_n(x, z) \le d_n(x, y) + d_n(y, z) \implies 2^n \le 1 + 1 = 2 \]

Since we are in the case of \(n \ge 2\) (and \(n\) is an integer), \(2^n \ge 4 > 2\). This clearly contradicts the triangle inequality! Therefore, when \(n \ge 2\), \(d_n\) cannot constitute a metric.

In conclusion, the only \(n\) satisfying the condition is \(1\).

Problem 9

Let \((S, d)\) be a metric space. Are we always able to define a metric \(d'\) on the power set \(\mathcal{P}(S)\) of \(S\), such that for any \(x, y \in S\), we have \(d'(\{x\}, \{y\}) = d(x, y)\)? Please prove or provide your reasoning.

Solution

Conclusion: Yes, we can always construct such a metric \(d'\).

Construction and Proof: The core of this problem lies in: viewing singletons as copies of \(S\), how can we legitimately "extend" the metric defined on the subset of singletons to the entire power set \(\mathcal{P}(S)\) containing non-singletons, without breaking the three metric axioms (especially the triangle inequality).

  • Trivial case: If \(S = \emptyset\), then \(\mathcal{P}(S) = \{\emptyset\}\). Defining the trivial metric \(d'(\emptyset, \emptyset) = 0\) on this singleton set is sufficient.

  • General case: Let \(S\) be non-empty. We fix any arbitrary point \(x_0 \in S\). For any two sets \(A\) and \(B\) in \(\mathcal{P}(S)\), we define the metric \(d'(A, B)\) as follows:

  • If \(A = B\), define \(d'(A, B) = 0\).

  • If \(A = \{x\}\) and \(B = \{y\}\) (i.e., both are singletons, and \(x \neq y\)), define \(d'(A, B) = d(x, y)\).

  • If \(A = \{x\}\) is a singleton, and \(B\) is not a singleton, define \(d'(A, B) = d(x, x_0) + 1\). By symmetry, if \(A\) is not a singleton and \(B=\{y\}\), then define \(d'(B, A) = d'(A, B) = d(y, x_0) + 1\).

  • If both \(A\) and \(B\) are not singletons (and \(A \neq B\)), define \(d'(A, B) = 2\).

Next, we verify that this is indeed a valid metric:

  • Strict Positivity and Symmetry: Because \(d(x,y)>0\) (when \(x \neq y\)), and \(d(x, x_0)+1 \ge 1 > 0\) as well as \(2 > 0\), we have \(d'(A,B)=0\) if and only if \(A=B\). Symmetry is directly guaranteed by definition.

  • Verification of the Triangle Inequality \(d'(A, C) \le d'(A, B) + d'(B, C)\) (assuming \(A, B, C\) are pairwise distinct):

  • All are singletons (\(A=\{x\}, B=\{y\}, C=\{z\}\)): \(d(x,z) \le d(x,y) + d(y,z)\), guaranteed by the triangle inequality of the original metric \(d\).

  • Two singletons, one non-singleton (\(A=\{x\}, B=\{y\}\), \(C\) non-singleton): We need to prove \(d'(A,C) \le d'(A,B) + d'(B,C)\). That is, to prove \(d(x, x_0) + 1 \le d(x, y) + d(y, x_0) + 1\). Canceling the 1 yields \(d(x, x_0) \le d(x, y) + d(y, x_0)\), which is guaranteed by the original metric \(d\) and thus holds. Similarly, we need to prove \(d'(A,B) \le d'(A,C) + d'(C,B)\). That is, to prove \(d(x,y) \le d(x, x_0) + 1 + d(y, x_0) + 1\). This obviously holds because \(d(x,y) \le d(x, x_0) + d(y, x_0) \le d(x, x_0) + d(y, x_0) + 2\).

  • One singleton, two non-singletons (\(A=\{x\}\), \(B\) non-singleton, \(C\) non-singleton): We need to prove \(d'(A,C) \le d'(A,B) + d'(B,C)\). That is, to prove \(d(x, x_0) + 1 \le d(x, x_0) + 1 + 2\), which obviously holds. Similarly, we need to prove \(d'(B,C) \le d'(B,A) + d'(A,C)\). That is, to prove \(2 \le d(x, x_0) + 1 + d(x, x_0) + 1 = 2d(x, x_0) + 2\). Since distance is non-negative (\(d(x,x_0) \ge 0\)), the right side of the inequality is at least 2, which clearly holds.

  • All non-singletons: We need to prove \(2 \le 2 + 2\), which obviously holds.

All metric axioms are satisfied, and according to the definition, when \(A=\{x\}, B=\{y\}\), \(d'(A,B) = d(x,y)\). Therefore, such a metric \(d'\) always exists.

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