Yes, the statement is true. Here's an explanation:

Definitions:

1. $A \subseteq \mathbb{R}$: $A$ is a subset of the real numbers.
2. Bounded Above: A set $A$ is bounded above if there exists some real number $M$ such that $a \leq M$ for all $a \in A$. The least such $M$ is called the supremum (or least upper bound), denoted by $\sup A$.
3. $x + A$: This represents the set $\{x + a : a \in A\}$, which means adding $x$ to each element of $A$.

Proof Outline:

Given that $A$ is bounded above, there exists $M \in \mathbb{R}$ such that $a \leq M$ for all $a \in A$. This implies that $\sup A$ exists and is finite.

1. Bounded Above:

   • For any $a \in A$, $x + a \leq x + M$. Thus, $x + M$ is an upper bound for the set $x + A$.
   • Therefore, $x + A$ is bounded above by $x + M$.

2. Supremum:

   • We want to show that $\sup(x + A) = x + \sup A$.
   • Upper Bound: Let $s = \sup A$. Then for any $a \in A$, $a \leq s$, so $x + a \leq x + s$. Hence, $x + s$ is an upper bound for $x + A$.
   • Least Upper Bound: Suppose $y$ is any upper bound for $x + A$. Then for all $a \in A$, $x + a \leq y$. This implies $a \leq y - x$ for all $a \in A$, so $y - x$ is an upper bound for $A$. Therefore, $\sup A \leq y - x$, or equivalently, $x + \sup A \leq y$.
   • This shows that $x + \sup A$ is the least upper bound for $x + A$, i.e., $\sup(x + A) = x + \sup A$.

Conclusion:

The set $x + A$ is indeed bounded above, and $\sup(x + A) = x + \sup A$.

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Would you like more details or have any questions?

Here are some related questions for further exploration:

1. What happens if $A$ is not bounded above?
2. How does the result change if $A$ is bounded below instead of above?
3. Can you prove a similar result for the infimum of a set?
4. How does this result apply to sequences?
5. How would this proof change if $A$ were a finite set?

Tip: Understanding the properties of suprema and infima is crucial when dealing with bounded sets in real analysis.