# Let $R=[p_1,q_1]\times \cdots\times[p_n,q_n]$ and show that diam $R=d(p,q)=[\sum_{k=1}^n (q_k-p_k)^2]^{1\over 2}$.

Let $p=(p_1,p_2,...,p_n)$ and $q=(q_1,q_2,...,q_n)$ be points in $\mathbb{R^n}$ with $p_k<q_k$ for each $k$. Let $R=[p_1,q_1]\times \cdots\times[p_n,q_n]$ and show that diam $R=d(p,q)=[\sum_{k=1}^n (q_k-p_k)^2]^{1\over 2}$.

We're given two sequences in $\mathbb{R^n}$ with $p_k<q_k$ for each $k$. We let $R$ be the Cartesian product of the two sequences. I'm not sure how to go about showing this equality. Any solutions/hints are greatly appreciated.

• Pythagorean theorem. – T. Bongers Jan 18 '16 at 15:44
• How can I use that? – Happy Jan 19 '16 at 2:48
• Can you draw the picture in $\mathbb{R}^2$? – T. Bongers Jan 19 '16 at 2:58
• Yeah, I see how it works in 2d and 3d, I just don't know how to do it in general. – Happy Jan 19 '16 at 4:06

Given two points $x=(x_1,x_2,\ldots,x_n)$ and $y=(y_1,y_2,\ldots,y_n)$ in $R$ we have $p_k\le x_k\le q_k$ and $p_k\le y_k\le q_k$ for $k=1,2,\ldots,n$, then $$-(q_k-p_k)\le y_k-x_k\le q_k-p_k\quad\text{then}\quad 0\le \sum_{k=1}^n(y_k-x_k)^2\le \sum_{k=1}^n(q_k-p_k)^2$$ So $$d(x,y)\le d(p,q)$$