Math Problem Statement

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

Solution

To prove that the set S={(x,y)x2+y24}S = \{(x, y) \mid x^2 + y^2 \leq 4 \} is convex, we need to show that for any two points A=(x1,y1)A = (x_1, y_1) and B=(x2,y2)B = (x_2, y_2) in the set SS, every point on the line segment joining AA and BB is also in the set SS.

Step-by-Step Proof:

  1. Restating the Problem:
    The set SS is the interior (including the boundary) of a circle with radius 2 centered at the origin. This means that for any point (x,y)S(x, y) \in S, it holds that x2+y24x^2 + y^2 \leq 4.

  2. Convexity Definition:
    A set SS is convex if, for any two points A=(x1,y1)SA = (x_1, y_1) \in S and B=(x2,y2)SB = (x_2, y_2) \in S, the line segment joining AA and BB, defined by the parametric equation
    P(t)=(1t)A+tB=((1t)x1+tx2,(1t)y1+ty2)P(t) = (1 - t) A + t B = ((1 - t) x_1 + t x_2, (1 - t) y_1 + t y_2) for t[0,1]t \in [0, 1], is also contained in SS.

  3. Parametric Point on the Line Segment:
    Consider the point on the line segment P(t)=(x(t),y(t))=((1t)x1+tx2,(1t)y1+ty2)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 SS:
    To check if the point P(t)P(t) lies in SS, we need to verify whether it satisfies the condition for being in SS, which is
    x(t)2+y(t)24.x(t)^2 + y(t)^2 \leq 4.

    Let's compute x(t)2+y(t)2x(t)^2 + y(t)^2: x(t)=(1t)x1+tx2,y(t)=(1t)y1+ty2,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=((1t)x1+tx2)2+((1t)y1+ty2)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=(1t)2(x12+y12)+2t(1t)(x1x2+y1y2)+t2(x22+y22).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 AA and BB:
    Since A=(x1,y1)SA = (x_1, y_1) \in S and B=(x2,y2)SB = (x_2, y_2) \in S, we know that x12+y124x_1^2 + y_1^2 \leq 4 and x22+y224x_2^2 + y_2^2 \leq 4. Therefore, we can bound the above expression as follows: x(t)2+y(t)2(1t)24+2t(1t)(x1x2+y1y2)+t24.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)24((1t)2+t2)+2t(1t)(x1x2+y1y2).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 (1t)2+t2=12t+t2+t2=1(1 - t)^2 + t^2 = 1 - 2t + t^2 + t^2 = 1, so we get: x(t)2+y(t)24+2t(1t)(x1x2+y1y2).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 x12+y124x_1^2 + y_1^2 \leq 4 and x22+y224x_2^2 + y_2^2 \leq 4, implying that x(t)2+y(t)24x(t)^2 + y(t)^2 \leq 4.

Conclusion:

Since x(t)2+y(t)24x(t)^2 + y(t)^2 \leq 4 for all t[0,1]t \in [0, 1], every point on the line segment between AA and BB remains inside the set SS. Thus, SS is convex.


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)x2+y2=4}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.

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Math Problem Analysis

Mathematical Concepts

Convexity
Geometry
Algebra

Formulas

x² + y² ≤ 4
Parametric equation for a line segment: P(t) = (1 - t)A + tB

Theorems

Definition of Convex Sets

Suitable Grade Level

College Level (Advanced Geometry and Linear Algebra)