(2)Let $$ S \subseteq \mathbb{R} $$ and supphse that $$ s^{*}=\sup S $$ belongs to $$ S $$. If $$ u
eq S $$, show that $$ \sup (S \cup\{u\})=\sup \left\{s^{*}, u ight\} $$.

Question

(2)Let $$ S \subseteq \mathbb{R} $$ and supphse that $$ s^{*}=\sup S $$ belongs to $$ S $$. If $$ u 
eq S $$, show that $$ \sup (S \cup\{u\})=\sup \left\{s^{*}, uight\} $$.

UpStudy AI · Accepted Answer

To show that $$ \sup (S \cup \{u\}) = \sup \{s^{*}, u\} $$, we will follow these steps:

1. **Understanding the Definitions**:
   - Let $$ s^{*} = \sup S $$. This means that $$ s^{*} $$ is the least upper bound of the set $$ S $$.
   - The set $$ S \cup \{u\} $$ is the union of $$ S $$ and the singleton set containing $$ u $$.
   - We need to show that the supremum of the union is equal to the supremum of the set containing $$ s^{*} $$ and $$ u $$.

2. **Establishing Upper Bounds**:
   - Since $$ s^{*} $$ is an upper bound for $$ S $$, it follows that $$ s^{*} \geq s $$ for all $$ s \in S $$.
   - Therefore, $$ s^{*} $$ is also an upper bound for $$ S \cup \{u\} $$ if $$ u \leq s^{*} $$.
   - If $$ u > s^{*} $$, then $$ u $$ is an upper bound for $$ S \cup \{u\} $$.

3. **Finding the Least Upper Bound**:
   - We need to consider two cases based on the relationship between $$ u $$ and $$ s^{*} $$.

**Case 1**: $$ u \leq s^{*} $$
   - In this case, the upper bounds of $$ S \cup \{u\} $$ are $$ s^{*} $$ and $$ u $$, but since $$ s^{*} $$ is greater than or equal to $$ u $$, the least upper bound is $$ s^{*} $$.
   - Thus, $$ \sup (S \cup \{u\}) = s^{*} = \sup \{s^{*}, u\} $$.

**Case 2**: $$ u > s^{*} $$
   - Here, $$ u $$ is greater than all elements in $$ S $$ (since $$ s^{*} $$ is the least upper bound).
   - Therefore, $$ u $$ is an upper bound for $$ S \cup \{u\} $$.
   - The least upper bound in this case is $$ u $$ itself, since $$ s^{*} $$ is not an upper bound for $$ u $$.
   - Thus, $$ \sup (S \cup \{u\}) = u = \sup \{s^{*}, u\} $$.

4. **Conclusion**:
   - In both cases, we have shown that $$ \sup (S \cup \{u\}) = \sup \{s^{*}, u\} $$.
   - Therefore, we conclude that if $$ s^{*} = \sup S $$ belongs to $$ S $$ and $$ u 
eq S $$, then $$ \sup (S \cup \{u\}) = \sup \{s^{*}, u\} $$.

This completes the proof.
If $$ s^{*} $$ is the supremum of $$ S $$ and $$ u $$ is not in $$ S $$, then the supremum of $$ S \cup \{u\} $$ is the greater of $$ s^{*} $$ and $$ u $$.

(2)Let \( S \subseteq \mathbb{R} \) and supphse that \( s^{}=\sup S \) belongs to \( S \). If \( u \neq S \), show that \( \sup (S \cup\{u\})=\sup \left\{s^{}, u\right\} \).

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(2)Let \( S \subseteq \mathbb{R} \) and supphse that \( s^{*}=\sup S \) belongs to \( S \). If \( u \neq S \), show that \( \sup (S \cup\{u\})=\sup \left\{s^{*}, u\right\} \).