Certainly! Let's break down and explain the statement step by step: --- **Step 1: Understanding the context and terms** - The lemma is about a subset \( S \) of a metric space \((X, d)\), where \( d \) is the metric (distance function). - The lemma lists "equivalent" conditions (though only (b) is visible). - Condition (b) is: "For every \( \varepsilon > 0 \), there exists a finite \( T \subseteq X \) such that for each \( x \in S \), there is a \( y \in T \) with \( d(x, y) < \varepsilon \)." **Explanation:** This says that for any chosen "radius" \( \varepsilon \), you can find a finite set of points in \( X \) (called \( T \)) so that every point in \( S \) is within distance \( \varepsilon \) of some point in \( T \). --- **Step 2: Interpreting the meaning of condition (b)** - This is the definition of "total boundedness" for the set \( S \) in the metric space \((X, d)\). - In other words, \( S \) is called totally bounded if for every \( \varepsilon > 0 \), you can cover \( S \) with finitely many open balls of radius \( \varepsilon \) (centered at points in \( T \)). **Explanation:** Total boundedness is a stronger property than boundedness. It means you can always "approximate" all points in \( S \) by finitely many points (the centers of the covering balls) as closely as you like, depending on \( \varepsilon \). --- **Final Answer:** Condition (b) states that for every \( \varepsilon > 0 \), there exists a finite set \( T \subseteq X \) such that every point in \( S \) is within \( \varepsilon \) of some point in \( T \). This is precisely the definition of "total boundedness" of the set \( S \) in the metric space \((X, d)\).