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

[robertdeniro]

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

 

[Dick]

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.

 

[robertdeniro]

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

 

[Dick]

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?

 

[robertdeniro]

What does the set (x+B(y,r))/2 look like?

the set also have to be contained in S

 

[Dick]

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?

 

[robertdeniro]

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...

 

[Dick]

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?

 

[robertdeniro]

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)?

 

[Dick]

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.

 

[robertdeniro]

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?