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Math Help - Fun easy problem

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Fun easy problem

    Here is a problem I came up with that is extremely easy, but when I posted it on another forum it took a while for someone to answer and the answer was much more complex than what was needed. Let's see if MHF has better mathematicians....I have faith!

    Question: Find the largest n in \mathbb{R}^n such that every k-cell (denoted E) has finitely many points p,q\in{E} such that d(q,p)=\text{diam}\left(E\right). Prove that this is the largest n.
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  2. #2
    Moo
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    A Cute Angle Moo's Avatar
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    What are diam and k-cell ?
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Moo View Post
    What are diam and k-cell ?
    In a metric space

    {\rm{diam}}(E)=\sup_{a,b \in E}d(a,b),

    that is it is the diameter of the set E (where d(a,b) is the "distance from a to b, that is d is the metric).

    Now my guess is that k and n are supposed to be the same thing, but who am I to say.

    What supprises me is that with the level of clarity of the statement of this problem Mathstud28 got any solutions let alone one correct but more complicated ( ) than what he thought was needed.

    CB
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    What are diam and k-cell ?
    In my books they don't say n-cell...hmm...maybe they do. but anyways CaptainBlack said what \text{diam}(E), and a k-cell is defined to be: if a_i<b_i~~{i=1,\cdots{k}} then a k-cell is the set of all points \bold{p}=(p_1,\cdots{p_k}) in \mathbb{R}^k whose coordinates satisfy the inequalities a_i\leqslant{a_i}\leqslant{b_i}~~1\leqslant{i}\leq  slant{k}
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