I'm trying to understand Theorem 2.40 in Rudin's Principles of Mathematical Analysis, which states that every $k$-cell is compact. I'm stuck at the very first step which is defining the "maximal" distance in the $k$-cell.
Rudin lets $I$ be a $k$-cell consisting of points $x = (x_1, \ldots, x_n)$ where. $a_j \leq x_j \leq b_j$ for each $j$. He then defines
$$
\delta = \left(\sum\limits_{j=1}^k (b_j – a_j)^2\right)^{1/2},
$$
and claims that for any $x,y \in I$, we have $|x-y| \leq \delta$. I'm not sure why this is the case. I believe $\delta$ is the diameter of $I$. Attempting to expand this out:
\begin{align*}
\text{diam}(I) &= \sup\limits_{x,y \in I} |x-y| \\
&= \sup\limits_{x,y \in I} \left(\sum\limits_{j=1}^n (x_i – y_i)^2\right)^{1/2}.
\end{align*}
Because the square root function is increaasing, the problem is reduced to showing that $$
\sum\limits_{j=1}^n (x_i – y_i)^2 \leq \sum\limits_{j=1}^n (b_j – a_j)^2.
$$
I can attempt to compare term by term. I know that $a_j \leq x_j \leq b_j$ and $a_j \leq y_j \leq b_j$. I can't subtract this inequalities, because then I'd get $0 \leq x_j – y_j \leq 0$, which doesn't make any sense. I can try to work with the absolute value of the difference. But then, because $x_i \geq a_i$ and $-y_i \geq -b_i$, I get
$$
|x_i – y_i| \geq x_i – y_i \geq a_i – y_i \geq a_i – b_i,
$$
which is the opposite of what I'm trying to prove.
I'd appreciate any help in parsing this.
Best Answer
We have $a_j \leq x_j \leq b_j$ and $a_j \leq y_j \leq b_j$, multiplying the second inequality with $-1$ yields $-b_j \leq -y_j \leq- a_j$. Finally adding the two: $$ a_j - b_j \leq x_j - y_j \leq b_j - a_j,$$ from which the rest follows. So the only problem in your solution is the error when "substracting" the inequalities.