NO
DATE:
2)a) Let \( A, B \subset \mathbb{R} \) be non-empty subsets that are bounded above.
Set \( A+B \) is defined as: \( A+B=\{a+b: a \in A, b \in B\} \)
Definition: Theorem 4.2 (i) Let \( \delta \in \mathbb{R} \) be an upper bound for a set \( A C R \).
Then \( \delta=\sup (A) \Leftrightarrow \forall \varepsilon>0 \exists \) an \( a \in A \) s.t \( \delta-\varepsilon<a \).
Prove: \( \sup (A+B)=\sup (A)+\sup (B) \)
Let \( \delta=\sup (A) \) and \( B=\sup (B) \).
Prove two inequalities which are \( \sup (A+B) \leq \delta+\beta, \sup (A+B) \geq \delta+\beta \)
\[
\sup (A+B) \leqslant \delta+B
\]
\( \because \delta=\sup (A), \forall a \in A, a \leq \delta . \forall b \in B, b \leq \beta \).
\( \therefore \forall a \in A \) and \( \forall b \in B: \quad a+b \leq \delta+\beta \)
Hence, \( \forall A+B \leq \delta+\beta \Rightarrow \sup (A+B) \leq \delta+\beta \)
\[
\begin{array}{l}
\sup (A+B) \geq \delta+\beta \\
\because \delta=\sup (A), \forall \varepsilon>0, \exists a_{\varepsilon} \in A \text { s.t } \delta-\varepsilon<a_{\varepsilon} \leq \delta \\
\because \beta=\sup (B), \forall \varepsilon>0, \exists b_{\varepsilon} \in B \text { s.t } \beta-\varepsilon<b_{\varepsilon} \leq \beta
\end{array}
\]
Therefore,
\( a_{\varepsilon}+b_{\varepsilon}>\delta+\beta-2 \varepsilon \). As \( \varepsilon \) approaches zero,
\( a_{\varepsilon}+b_{\varepsilon} \) approaches \( \delta+\beta \).
Thus, \( \sup (A+B) \geq \delta+\beta \).
Since \( \sup (A+B) \leq \delta+\beta \) and \( \sup (A+B) \geq \delta+B \), it follows that \( \sup (A+B)=\delta+\beta=\sup (A)+\sup (B) \)
