Consider the budget set
(p)≡©x∈R+ :p·x≤a
with price vector p ∈ R++ (i.e.,   0 for every ) and income   0 Show that
(p) is a compact set.


To show that the budget set $ \mathcal{B}(p, I) \equiv \{ x \in \mathbb{R}^n_+ : p \cdot x \leq I \} $ is a compact set, we need to prove two things:

1. **The set is bounded**: A set is bounded if there exists a real number $ M $ such that the norm of every element in the set is less than $ M $.
2. **The set is closed**: A set is closed if it contains all its limit points, i.e., if a sequence of points within the set converges, its limit is also within the set.

### 1. Boundedness

Consider the set $ \mathcal{B}(p, I) \equiv \{ x \in \mathbb{R}^n_+ : p \cdot x \leq I \} $, where $ p $ is the price vector $ p = (p_1, p_2, \dots, p_n) $ and $ x = (x_1, x_2, \dots, x_n) $ is the vector of goods. The condition $ p \cdot x \leq I $ can be written as:
\[
p_1x_1 + p_2x_2 + \dots + p_nx_n \leq I.
\]

Since $ p_i > 0 $ for all $ i $, if any $ x_i $ were to grow without bound (i.e., $ x_i \to \infty $), the left-hand side $ p_1x_1 + \dots + p_nx_n $ would exceed $ I $, violating the condition $ p \cdot x \leq I $. Therefore, the components $ x_i $ cannot grow without bound. This implies that each $ x_i $ must be less than or equal to $ \frac{I}{p_i} $, establishing an upper bound for each component of $ x $. Thus, the set $ \mathcal{B}(p, I) $ is bounded.

### 2. Closedness

To show that $ \mathcal{B}(p, I) $ is closed, consider a convergent sequence $ \{x^k\} $ within the set $ \mathcal{B}(p, I) $, such that $ x^k \to x $ as $ k \to \infty $. Since the set $ \mathcal{B}(p, I) $ is defined by the inequality $ p \cdot x^k \leq I $, taking the limit as $ k \to \infty $ gives:
\[
p \cdot x = \lim_{k \to \infty} p \cdot x^k \leq I.
\]
Thus, the limit $ x $ also satisfies the condition $ p \cdot x \leq I $, meaning $ x \in \mathcal{B}(p, I) $. Therefore, $ \mathcal{B}(p, I) $ is closed.

### Conclusion: Compactness

A set in $ \mathbb{R}^n $ is compact if it is both closed and bounded. Since we have shown that $ \mathcal{B}(p, I) $ is both closed and bounded, it follows that $ \mathcal{B}(p, I) $ is compact.

Would you like further details or have any questions?

Here are five related questions you might find interesting:

1. How does the budget set change if one of the prices in the vector $ p $ is zero?
2. What happens to the compactness of the budget set if income $ I $ becomes unbounded?
3. Can you describe the geometrical interpretation of the budget set in $ \mathbb{R}^2 $?
4. How would introducing a minimum consumption requirement $ x_i \geq m_i > 0 $ for each good $ i $ affect the budget set?
5. What are the implications of compactness for optimization problems, such as utility maximization?

**Tip**: Compactness is crucial in optimization problems because it ensures the existence of an optimal solution within the budget set.

Explore the compactness of budget sets in economic theory with a detailed analysis. Understand how the budget set $\mathcal{B}(p, I)$, defined by the price vector $p$ and income $I$, is both bounded and closed, ensuring its compactness in mathematical terms.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation