The statement you’ve written seems to be a proposition involving two sets, $A$ and $B$, where $A \subseteq B$ and we are discussing the boundedness and supremum (least upper bound) of these sets.

Rephrased statement:

• If $B$ is bounded above, then $A$ is also bounded above.
• Moreover, the supremum of $A$ is less than or equal to the supremum of $B$.

Proof Outline:

1. Boundedness of $A$:
   If $B$ is bounded above, it means there exists some $M \in \mathbb{R}$ such that $b \leq M$ for all $b \in B$. Since $A \subseteq B$, every element of $A$ is also an element of $B$. Thus, $a \leq M$ for all $a \in A$. Therefore, $A$ is also bounded above by $M$.

2. Relationship of Supremums:
   Since $A \subseteq B$, the least upper bound (supremum) of $A$, denoted $\sup A$, cannot be greater than the least upper bound of $B$, denoted $\sup B$. Formally: $\sup A \leq \sup B$ This holds because any upper bound of $B$ is also an upper bound of $A$ (since $A \subseteq B$), and $\sup B$ is the least such upper bound.

Conclusion:

• $B$ being bounded above implies $A$ is also bounded above.
• The supremum of $A$ is less than or equal to the supremum of $B$, i.e., $\sup A \leq \sup B$.

Would you like further details or any clarifications on this?

Related Questions:

1. What is the difference between supremum and maximum?
2. Can the supremum of a set be an element of the set?
3. What happens if $A$ and $B$ are equal? What can be said about their supremums?
4. How do we formally define a bounded set in real analysis?
5. What is the significance of the least upper bound property in the real numbers?

Tip:

Always remember that the supremum of a set may not be in the set itself, but it is always the smallest possible upper bound.