How do you prove that the set S= {(x,y) |x²+y²|<=4} is a convex set?


To prove that the set \( S = \{(x, y) \mid x^2 + y^2 \leq 4 \} \) is convex, we need to show that for any two points \( A = (x_1, y_1) \) and \( B = (x_2, y_2) \) in the set \( S \), every point on the line segment joining \( A \) and \( B \) is also in the set \( S \).

### Step-by-Step Proof:

1. **Restating the Problem**:  
   The set \( S \) is the interior (including the boundary) of a circle with radius 2 centered at the origin. This means that for any point \( (x, y) \in S \), it holds that \( x^2 + y^2 \leq 4 \).

2. **Convexity Definition**:  
   A set \( S \) is convex if, for any two points \( A = (x_1, y_1) \in S \) and \( B = (x_2, y_2) \in S \), the line segment joining \( A \) and \( B \), defined by the parametric equation  
   \[
   P(t) = (1 - t) A + t B = ((1 - t) x_1 + t x_2, (1 - t) y_1 + t y_2)
   \]
   for \( t \in [0, 1] \), is also contained in \( S \).

3. **Parametric Point on the Line Segment**:  
   Consider the point on the line segment \( P(t) = (x(t), y(t)) = ((1 - t) x_1 + t x_2, (1 - t) y_1 + t y_2) \).

4. **Check if Point Lies in \( S \)**:  
   To check if the point \( P(t) \) lies in \( S \), we need to verify whether it satisfies the condition for being in \( S \), which is  
   \[
   x(t)^2 + y(t)^2 \leq 4.
   \]

   Let's compute \( x(t)^2 + y(t)^2 \):
   \[
   x(t) = (1 - t) x_1 + t x_2, \quad y(t) = (1 - t) y_1 + t y_2,
   \]
   \[
   x(t)^2 + y(t)^2 = \left( (1 - t) x_1 + t x_2 \right)^2 + \left( (1 - t) y_1 + t y_2 \right)^2.
   \]

   By expanding the terms, we get:
   \[
   x(t)^2 + y(t)^2 = (1 - t)^2 (x_1^2 + y_1^2) + 2t(1 - t)(x_1 x_2 + y_1 y_2) + t^2 (x_2^2 + y_2^2).
   \]

5. **Use the Condition for \( A \) and \( B \)**:  
   Since \( A = (x_1, y_1) \in S \) and \( B = (x_2, y_2) \in S \), we know that \( x_1^2 + y_1^2 \leq 4 \) and \( x_2^2 + y_2^2 \leq 4 \). Therefore, we can bound the above expression as follows:
   \[
   x(t)^2 + y(t)^2 \leq (1 - t)^2 \cdot 4 + 2t(1 - t)(x_1 x_2 + y_1 y_2) + t^2 \cdot 4.
   \]
   This simplifies to:
   \[
   x(t)^2 + y(t)^2 \leq 4 \left( (1 - t)^2 + t^2 \right) + 2t(1 - t)(x_1 x_2 + y_1 y_2).
   \]

6. **Final Verification**:  
   The term \( (1 - t)^2 + t^2 = 1 - 2t + t^2 + t^2 = 1 \), so we get:
   \[
   x(t)^2 + y(t)^2 \leq 4 + 2t(1 - t)(x_1 x_2 + y_1 y_2).
   \]
   The second term does not exceed 4 because \( x_1^2 + y_1^2 \leq 4 \) and \( x_2^2 + y_2^2 \leq 4 \), implying that \( x(t)^2 + y(t)^2 \leq 4 \).

### Conclusion:
Since \( x(t)^2 + y(t)^2 \leq 4 \) for all \( t \in [0, 1] \), every point on the line segment between \( A \) and \( B \) remains inside the set \( S \). Thus, \( S \) is convex.

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Do you have any questions or would you like further details on any part of the proof?

Here are 5 related questions:
1. What is the geometric interpretation of convexity in higher dimensions?
2. How can you prove convexity for more complex sets, like ellipses or polygons?
3. Can a set that contains holes or gaps still be convex?
4. How does convexity relate to optimization problems in mathematics?
5. How would you prove that the set \( S = \{(x, y) \mid x^2 + y^2 = 4 \} \) (the boundary of the circle) is not convex?

**Tip**: When proving convexity, checking line segments between points within the set is a fundamental method.

How do you prove that the set S = {(x, y) | x² + y² ≤ 4} is a convex set?

This detailed step-by-step proof demonstrates how to prove that the set S = {(x, y) | x² + y² ≤ 4}, the interior of a circle, is a convex set. Learn key concepts such as convexity, parametric equations, and geometric proof techniques. Suitable for college-level students studying advanced geometry or linear algebra.

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