Question: All questions in this exam are compulsory. Question 1 Metric Spaces (20 marks) a. Define the…

All questions in this exam are compulsory.

Question 1 Metric Spaces (20 marks)

a. Define the following terms as defined in this course;

i) Let X be a non-empty set and d a metric on X.

a) Define the underline phrase (4 marks)

b) Let Y be a non-empty subset of X denoted Y ⊆ X. Write down the map defined on Y for which the pair (Y and the map) will be a metric space. (4 marks)

ii) Let X be a set. Define the following

i. A neighbourhood (1 mark) ii. Open sets (1 mark) iii. Closed sets (1 mark) iv. Limit point (2 marks) v. Closure (2 marks) vi. Connectedness of a metric space (2 marks)

b) Prove that given two metric spaces (X, d) and (Y, d|Y) and f: (X, d) → (Y, d|Y) a continuous map. X connected implies f(X) is as well connected (3 marks)

Solution

Question 1: Metric Spaces Part (a): Definitions i) Definition of a Metric Space A metric space is defined by a pair \((X, d)\), where: \[ X \text{ is a non-empty set} \] \[ d: X \times X \rightarrow \mathbb{R} \text{ is a metric on } X \] Part (b): Induced Metric on a Subset Given \( Y \subseteq X \), the metric \( d|_Y \) is defined as: \[ d|_Y: Y \times Y \rightarrow \mathbb{R} \] \[ d|_Y(y_1, y_2) = d(y_1, y_2) \quad \text{for all } y_1, y_2 \in Y \] This makes \((Y, d|_Y)\) a metric space. Part (c): Key Definitions in Metric Spaces i. Neighborhood A neighborhood of a point \( x \in X \) is an open set \( U \) containing \( x \). ii. Open Sets A set \( U \subseteq X \) is open if for every \( x \in U \), there exists an \( \epsilon > 0 \) such that: \[ B(x, \epsilon) = \{ y \in X \mid d(x, y) < \epsilon \} \subseteq U \] iii. Closed Sets A set \( C \subseteq X \) is closed if its complement \( X \setminus C \) is open. iv. Limit Point A point \( x \in X \) is a limit point of a set \( A \subseteq X \) if every neighborhood of \( x \) contains at least one point of \( A \) distinct from \( x \). v. Closure The closure of a set \( A \subseteq X \), denoted by \( \overline{A} \), is the smallest closed set containing \( A \). It includes all points of \( A \) and all limit points of \( A \). vi. Connectedness of a Metric Space A metric space \( (X, d) \) is connected if it cannot be partitioned into two non-empty disjoint open sets. In other words, there do not exist open sets \( U \) and \( V \) such that: \[ X = U \cup V \] \[ U \cap V = \emptyset \] and both \( U \) and \( V \) are non-empty. Part (d): Proof of Connectedness under Continuous Mapping Statement: Let \( f: (X, d) \rightarrow (Y, d|_Y) \) be a continuous map between metric spaces. If \( X \) is connected, then \( f(X) \) is connected. Proof: Assume, for contradiction, that \( f(X) \) is not connected. Then there exist non-empty disjoint open sets \( U \) and \( V \) in \( Y \) such that: \[ f(X) \subseteq U \cup V \] \[ f(X) \cap U \neq \emptyset \quad \text{and} \quad f(X) \cap V \neq \emptyset \] Define the preimages: \[ A = f^{-1}(U) \quad \text{and} \quad B = f^{-1}(V) \] Since \( f \) is continuous, \( A \) and \( B \) are open in \( X \). Also: \[ A \cup B = X \] \[ A \cap B = f^{-1}(U) \cap f^{-1}(V) = f^{-1}(U \cap V) = f^{-1}(\emptyset) = \emptyset \] Moreover, both \( A \) and \( B \) are non-empty because \( f(X) \cap U \neq \emptyset \) and \( f(X) \cap V \neq \emptyset \). This partitions \( X \) into two non-empty disjoint open sets \( A \) and \( B \), contradicting the connectedness of \( X \). Therefore, \( f(X) \) must be connected.