Proving that (x+y)/2 Belongs to Interior of S in Vectorspace

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In summary, the problem is asking to prove that (x+y)/2, where x is in S and y is in the interior of S, is also in the interior of S. This can be proven by showing that the set (x+B(y,r))/2, where B(y,r) is an open ball around y, is contained in S. This can be deduced by considering the fact that B(y,r)/2=B(y/2,r/2) and using the property that (a+b)/2 belongs to S for a, b in S. Therefore, (x+y)/2 must also be in the interior of S.
  • #1
robertdeniro
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Homework Statement


let V be a vector space and S be a subset of V.
S has the property that for a, b belonging in S, (a+b)/2 also belongs to S.

given: x belongs to S and y belongs to interior of S.
prove: (x+y)/2 belongs to interior of S.


Homework Equations





The Attempt at a Solution

 
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  • #2
Where's your attempt at a solution? If y is in the interior of S what does that tell you about vectors 'near' y? Come on, try.
 
  • #3
i was thinking of trying to represent vectors near (x+y)/2 as the average of 2 vectors in in S, this way we can deduce that the vector are also in S. but I am not sure how to pick these 2 vectors wisely.

to answer your question. if y is in the interior of S, then vectors near y are also in S
 
  • #4
Ok. So there is an open ball around y of some radius r, B(y,r) such that every vector in B(y,r) is in S, that's what 'near' means right? What does the set (x+B(y,r))/2 look like?
 
  • #5
What does the set (x+B(y,r))/2 look like?

the set also have to be contained in S
 
  • #6
robertdeniro said:
What does the set (x+B(y,r))/2 look like?

the set also have to be contained in S

Ok, then is that a neighborhood of (x+y)/2? Would that put (x+y)/2 in the interior of S?
 
  • #7
Dick said:
Ok, then is that a neighborhood of (x+y)/2? Would that put (x+y)/2 in the interior of S?

i guess yes. not sure how the algebra would work here...
 
  • #8
robertdeniro said:
i guess yes. not sure how the algebra would work here...

It's a vector space. Isn't B(y,r)/2=B(y/2,r/2)? Think about how a ball is defined. What would x/2+B(y/2,r/2) be in terms of a ball?
 
  • #9
Dick said:
It's a vector space. Isn't B(y,r)/2=B(y/2,r/2)? Think about how a ball is defined. What would x/2+B(y/2,r/2) be in terms of a ball?

you lost me here. why is B(y,r)=B(y/2, r/2)?
 
  • #10
robertdeniro said:
you lost me here. why is B(y,r)=B(y/2, r/2)?

It isn't. B(y,r)/2=B(y/2,r/2). B(y,r) is the set of all vectors y+x where |x|<r, isn't it? What's B(y,r)/2? You are getting lost too easily. Try to visualize it.
 
  • #11
oops didnt see the /2

so B(y,r)/2 is multiplying every element in the ball by 1/2. so i see where the B(y/2, r/2) is coming from now.

so (x+B(y,r))/2 = x/2 + B(y/2, r/2) = B((x+y)/2, r/2)?

so this means B((x+y)/2, r/2) is also contained in S, hence (x+y)/2 is in the interior?
 
Last edited:

What is the definition of "interior" in a vector space?

In a vector space, the interior of a set is the largest open set contained in that set. It consists of all the elements that do not touch the boundary of the set.

What is the significance of proving that (x+y)/2 belongs to the interior of S in a vector space?

This proof shows that the average of two elements in a set S is also an element of the set, and it therefore helps to establish the convexity of the set S. Convexity is an important concept in vector spaces and is used in various mathematical applications.

How do you go about proving that (x+y)/2 belongs to the interior of S in a vector space?

One approach is to show that there exists an open ball around (x+y)/2 that is completely contained within the set S. This can be done by choosing a radius for the ball that is smaller than the distance between (x+y)/2 and the boundary of S. Additionally, it is important to use the properties of vector spaces, such as closure under addition and scalar multiplication, in the proof.

What are some examples of vector spaces where this proof can be applied?

This proof can be applied to any vector space that satisfies the properties of closure under addition and scalar multiplication. Some examples include Euclidean spaces, function spaces, and matrix spaces.

Are there any real-world applications of this proof?

Yes, there are many real-world applications of this proof, particularly in optimization and convex analysis. For example, it can be used to show the existence of a solution to a linear programming problem, or to prove the convexity of a utility function in economics.

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