If x∈R and A is bounded above, then x+A is bounded above and sup⁡(x+A)=x+sup⁡A.

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.

Explore the concept of supremum in real analysis and its application to bounded sets. Learn how to prove that if x is a real number and A is a bounded above set, then x + A is also bounded above, and sup(x + A) = x + sup(A).

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