Fun easy problem

• Nov 27th 2008, 08:55 PM
Mathstud28
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 $n$.
• Nov 27th 2008, 11:38 PM
Moo
What are diam and k-cell ?
• Nov 28th 2008, 12:02 AM
CaptainBlack
Quote:

Originally Posted by Moo
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 ( (Rofl) ) than what he thought was needed.

CB
• Nov 29th 2008, 11:59 AM
Mathstud28
Quote:

Originally Posted by Moo
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 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 $a_i\leqslant{a_i}\leqslant{b_i}~~1\leqslant{i}\leq slant{k}$