Prove the following is a convex set?

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Homework Help Overview

The problem involves proving that a set defined by linear inequalities and non-negativity constraints is convex. The set is expressed as F = {x ∈ R^n : Ax ≥ b; x ≥ 0}, where A and b are parameters that need clarification.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the definition of convexity and express uncertainty about how to apply it to the given set. Questions arise regarding the nature of the parameters A and b, particularly whether they are vectors or scalars, and how the inequalities should be interpreted.

Discussion Status

The discussion is ongoing, with participants seeking to clarify the definitions and relationships between the variables involved. There is a focus on understanding the implications of the inequalities and the non-negativity condition.

Contextual Notes

Participants note confusion regarding the interpretation of vector inequalities and the nature of the parameters A and b, which are described as arbitrary but require further specification for clarity.

ashina14
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Homework Statement


Prove that F = {x E R^n : Ax >/= b; x >/= 0} is a convex
set.

Yes x in non negative and A and b are any arbitrary


Homework Equations






The Attempt at a Solution



Well I know A set T is convex if x1, x2 E T implies that px1+(1-p)x2 E T for all 0 <= p <= 1.
I don't know how to use this information.
 
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ashina14 said:

Homework Statement


Prove that F = {x E R^n : Ax >/= b; x >/= 0} is a convex
set.

Yes x in non negative and A and b are any arbitrary

Are arbitrary what? What does it mean for ##x\in R^n## to satisfy ##x > 0##?
 
A and b are any arbitrary number in R.
All vectors in x E Rn are non-negative
 
ashina14 said:

Homework Statement


Prove that F = {x E R^n : Ax >/= b; x >/= 0} is a convex
set.

LCKurtz said:
Are arbitrary what? What does it mean for ##x\in R^n## to satisfy ##x > 0##?

ashina14 said:
A and b are any arbitrary number in R.
All vectors in x E Rn are non-negative

This doesn't make any sense to me. If x is a vector and A is a real number then Ax is a vector. If b is a real number you have the vector Ax > b, a real number. What does it mean for a vector to be greater than a number? It makes about as much sense to say an apple is greater than a bicycle.
 
I'm so sorry. b is a vector too. I don't understand this topic too well.
 
That doesn't help. What does it mean for one vector to be greater than another?
 

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