To prove that $b \in S$ for a non-empty bounded closed set $S \subset \mathbb{R}$, where $b = \sup(S)$ (the supremum of $S$), we proceed as follows:

Proof:

1. $b = \sup(S)$:
   By definition, $b = \sup(S)$ means that:

   • $b$ is an upper bound of $S$, i.e., $x \leq b$ for all $x \in S$.
   • For every $\epsilon > 0$, there exists some element $x_\epsilon \in S$ such that $b - \epsilon < x_\epsilon$. This property follows from the definition of the supremum as the least upper bound.

2. Suppose $b \notin S$ (for contradiction):
   Assume, for the sake of contradiction, that $b \notin S$. Since $S$ is closed, this would mean that $b$ is not a limit point of $S$ either.

3. $S$ is closed:
   Since $S$ is closed, it contains all its limit points. Therefore, if $b$ is a limit point of $S$, it must belong to $S$. However, we assumed that $b \notin S$, so $b$ cannot be a limit point of $S$.

4. Contradiction from the supremum definition:
   Given that $b = \sup(S)$, for every $\epsilon > 0$, there exists $x_\epsilon \in S$ such that $b - \epsilon < x_\epsilon \leq b$. As $\epsilon \to 0$, the values of $x_\epsilon$ get arbitrarily close to $b$. This implies that $b$ is a limit point of $S$. Since $S$ is closed, it must contain all its limit points, including $b$.

5. Conclusion:
   The assumption that $b \notin S$ leads to a contradiction. Therefore, $b \in S$.

Thus, we have proved that $b = \sup(S) \in S$ when $S$ is a non-empty bounded closed set in $\mathbb{R}$.

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Related Questions:

1. What is the definition of a supremum in general sets?
2. How does the property of closed sets ensure all limit points are included?
3. Can the result be generalized to unbounded sets in $\mathbb{R}$?
4. How does the concept of infimum relate to this proof?
5. What would happen if the set $S$ was not closed?

Tip: Always verify if a set is closed when working with suprema or infima to ensure limit points are considered properly.