Convexity prob: Prove that all boxes in R^m are convex

[passionate]

Can someone please give me some help on this problem?
Prove that all boxes in R^m are convex.

 

[pka]

Re: Convex problem

What is the exact definition of a box in \(\displaystyle R^n\)?

 

[galactus]

Re: Convex problem

I was wondering that also. I know the 'ball', but never heard of a 'box'.

 

[passionate]

Re: Convex problem

The problem itself doesn't mention anything regarding the definition of a box in R^m . I'm helpless

 

[stapel]

The problem itself doesn't mention anything regarding the definition of a box in R^m

The exercises generally depend upon definitions provided in the preceding section(s) of the book. If the book does not provide the definition, then you'll need to consult with your instructor regarding the missing information.

Thank you!

Eliz.

 

[passionate]

So, I've been reviewing some definitions and see that the section does mention what a box is. However, it doesn't give any formal definition. There is a picture in the section showing a box, which is a 3-dimensional rectangular box and a cube. That's all it says about "boxes". The rest is more on the unit ball.

 

[stapel]

...the section does mention what a box is. However, it doesn't give any formal definition.

Mathematics, and proofs in particular, depend on definitions. If you haven't been provided one, then there isn't much you can do. :shock:

Kindly please consult with your instructor regarding the missing information. Thank you!

Eliz.

 

[passionate]

I just got the definition of a box from my prof.
box : [a_1, b_1] x [a_2, b_2] x...x [a_m, b_m]

where x means times and 1,2,...,m are the subscripts.

 

[pka]

There really is nothing to prove here except to make of observations about the obvious. But because you seem unsure about the definitions, I will review some basics. To say that K is convex in \(\displaystyle R^n\) means that \(\displaystyle \left\{ {a,b} \right\} \subseteq K\quad \Rightarrow \quad \left\{ {ta + \left( {1 - t}\right)b \,:0 \leqslant t \leqslant 1} \right\}\subseteq K\).
Now the \(\displaystyle \times\) in the definition is not ‘times’ but cross product.
Thus each point in a box K is an ordered n-tuple.
So on the interval \(\displaystyle \left[ {x_j ,y_j } \right]\) the points on the ‘line’ defined above are \(\displaystyle \left\{ {tx_j + \left( {1 - t} \right)y_j : \, 0 \le t \le 1} \right\}\).
Now here is the obvious part: we know that intervals of real numbers are convex.
So each point ‘between’ two points of K is also a point of K.
That proves the proposition.
I hope that this has helped you. However, if you are still totally lost with this then I suggest you have a serious sit-down, one-on-one with your professor.

 

[passionate]

There really is nothing to prove here except to make of observations about the obvious. But because you seem unsure about the definitions, I will review some basics. To say that K is convex in \(\displaystyle R^n\) means that \(\displaystyle \left\{ {a,b} \right\} \subseteq K\quad \Rightarrow \quad \left\{ {ta + \left( {1 - t}\right)b \,:0 \leqslant t \leqslant 1} \right\}\subseteq K\).
Now the \(\displaystyle \times\) in the definition is not ‘times’ but cross product.
Thus each point in a box K is an ordered n-tuple.
So on the interval \(\displaystyle \left[ {x_j ,y_j } \right]\) the points on the ‘line’ defined above are \(\displaystyle \left\{ {tx_j + \left( {1 - t} \right)y_j : \, 0 \le t \le 1} \right\}\).
Now here is the obvious part: we know that intervals of real numbers are convex.
So each point ‘between’ two points of K is also a point of K.
That proves the proposition.
I hope that this has helped you. However, if you are still totally lost with this then I suggest you have a serious sit-down, one-on-one with your professor.

Thank you very much for your help. It really does help me understand convexity better. Geometrically, I see that a a box in R^m is convex since we can pick any two points belonging to the box, then connect the points as a line segment, we will see that the line segment is also in that box. However, I didn't know how to formally prove it based on the definition I have. The definition for convex sets I have is stated in a different way:
E in R^m is convex if for each x, y in E, the straight line segment between x and y is in E.

On the second line of your post, you used the notation {a,b} in K => {ta + (1 - t )b: where t is in [0,1] } is in K. What is {a,b} in your definition? Is it a set? and follows it is also a set? I have not seen this formula before, so I think I need to google and read more.
I greatly appreciate your help.

 

[pka]

On the second line of your post, you used the notation {a,b} in K => {ta + (1 - t )b: where t is in [0,1] } is in K. What is {a,b} in your definition? Is it a set? and follows it is also a set? I have not seen this formula before, so I think I need to google and read more.

I think you need to have that conference with the professor.

 

[daon]

{a,b} in K => {ta + (1 - t )b: where t is in [0,1] } is in K. What is {a,b} in your definition? Is it a set? and follows it is also a set? I have not seen this formula before, so I think I need to google and read more.
I greatly appreciate your help.

It means if a and b are any two points in a set K, the quantity p(t)=ta + (1-t)b is also a point in K for \(\displaystyle 0 \le t \le 1\). Its another way for saying that the straight-line path from b to a lies in the set K. Note p(t) can be written as (a-b)t + b (looks more like the equation of a line this way).