Sure! Let's **prove the equivalence** of the conditions in Lemma 02.1 in **two main steps**, explaining each step and then giving the final conclusion. --- ### **Step 1: (a) ⇒ (b) ⇒ (c)** #### (a) ⇒ (b): If \( S \) is totally bounded, then for every \( \varepsilon > 0 \), there exists a **finite \(\varepsilon\)-net** in \( S \). - This means: There is a finite subset \( T \subseteq S \) such that every \( x \in S \) is within \( \varepsilon \) of some \( y \in T \) (i.e., \( d(x, y) < \varepsilon \)). - Since \( T \subseteq S \subseteq X \), we can also view \( T \) as a subset of \( X \). - Therefore, **for every \( \varepsilon > 0 \), there exists a finite \( T \subseteq X \) such that for each \( x \in S \), there is \( y \in T \) with \( d(x, y) < \varepsilon \)**, which is exactly condition (b). **Explanation:** Total boundedness means you can "approximate" all points in \( S \) with finitely many points within any given accuracy. This gives you a finite \( \varepsilon \)-net as in (b). --- #### (b) ⇒ (c): Given (b), for every \( \varepsilon > 0 \), there is a finite \( T = \{y_1, \ldots, y_n\} \subseteq X \) such that for each \( x \in S \), \( d(x, y_i) < \varepsilon \) for some \( i \). - Consider the balls \( B(y_i, \varepsilon) \) for \( i = 1, ..., n \). - The union of these balls **covers \( S \)** (since every \( x \in S \) is close to some \( y_i \)). - Each ball has diameter at most \( 2\varepsilon \), but we can always adjust \( \varepsilon \) by using \( \varepsilon/2 \) instead. - **Thus:** For any \( \varepsilon > 0 \), \( S \) can be covered by finitely many subsets of \( X \) (the balls) with diameter less than \( \varepsilon \), which is (c). **Explanation:** Having a finite net implies you can cover the set with finitely many small balls, which is a finite cover with small diameter. --- ### **Step 2: (c) ⇒ (d) ⇒ (a)** #### (c) ⇒ (d): If for every \( \varepsilon > 0 \), there’s a finite cover of \( S \) by subsets of diameter less than \( \varepsilon \), - Then you can **refine the cover into a finite partition**: assign each point in \( S \) to one of the covering sets, forming a partition. - The resulting partition consists of finitely many sets, each of diameter less than \( \varepsilon \), which is (d). **Explanation:** A finite cover can always be refined to a finite partition by distributing points among the covering sets. --- #### (d) ⇒ (a): Suppose for every \( \varepsilon > 0 \), there is a finite partition of \( S \) into sets of diameter less than \( \varepsilon \). - Picking one point from each partition set gives a finite \( \varepsilon \)-net for \( S \). - Thus, \( S \) is totally bounded. **Explanation:** A finite partition with small diameter allows you to choose representatives forming a finite net, matching the definition of total boundedness. --- ## **Final Answer** **All four conditions (a)-(d) are equivalent** for a subset \( S \) of a metric space \((X, d)\): 1. **(a) ⇒ (b) ⇒ (c) ⇒ (d) ⇒ (a)** (as shown above). 2. **Explanation:** Each step shows how the property in one condition guarantees the next, eventually circling back to the first, thereby proving equivalence.

Certainly! Here's a clear two-step explanation with reasoning at each stage: --- ### **Step 1: Showing (a) ⇒ (b) and (b) ⇒ (c)** **(a) ⇒ (b):** - **Total boundedness** of \( S \) means: For every \( \varepsilon > 0 \), there exists a **finite \(\varepsilon\)-net** within \( S \). - **Implication:** We can find a finite subset \( T \subseteq S \) such that every point \( x \in S \) is within \( \varepsilon \) of some \( y \in T \). - **Result:** This matches condition (b): for each \( x \in S \), there exists \( y \in T \) with \( d(x, y) < \varepsilon \). **(b) ⇒ (c):** - Given (b), for each \( \varepsilon > 0 \), we have a finite set \( T = \{ y_1, y_2, \ldots, y_n \} \) such that every \( x \in S \) is within \( \varepsilon \) of some \( y_i \). - **Construct a cover:** Take balls \( B(y_i, \varepsilon) \). These finitely many balls cover \( S \). - **Diameter consideration:** Each ball has diameter at most \( 2\varepsilon \). By choosing smaller \( \varepsilon \), we can ensure the diameter of each covering set is less than \( \varepsilon \). - **Result:** \( S \) can be covered by finitely many subsets of diameter less than \( \varepsilon \), satisfying (c). --- ### **Step 2: Showing (c) ⇒ (d) and (d) ⇒ (a)** **(c) ⇒ (d):** - Given that for each \( \varepsilon > 0 \), \( S \) is covered by finitely many subsets each of diameter less than \( \varepsilon \). - **Refinement:** Partition \( S \) into these subsets, which gives a finite partition where each part has diameter less than \( \varepsilon \). - **Result:** Condition (d) holds. **(d) ⇒ (a):** - If for every \( \varepsilon > 0 \), \( S \) admits a finite partition into sets of diameter less than \( \varepsilon \), - **Choose representatives:** Pick one point from each part. - These points form a finite \( \varepsilon \)-net for \( S \). - **Conclusion:** \( S \) is totally bounded, satisfying (a). --- ### **Final conclusion:** **All four conditions are equivalent.** This means that **total boundedness** of \( S \) in \((X, d)\) can be characterized in multiple, equivalent ways as given in the lemma. --- **Answer:** **(a) \( S \) is totally bounded if and only if any of the conditions (b), (c), or (d) hold.**