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An intervall in $\mathbb{R}^n$ is a set of points $x = (x_1, \ldots, x_n)$ such that $$ a_i < x_i < b_i \qquad (i = 1, \ldots, n) $$ and where $<$ could also be replaced by $\le$. If $A$ is the union of a finite number of intervals, then $A$ is said to be an elementary set. Let $\mathscr{E}$ be the set of all elementary sets. Then every $A \in \mathscr{E}$ could be written as the union of a finite number of disjoint sets.

Okay, I want to proof that the set $B_r := \{ x : \sum_{i=1}^n x_i^2 < r^2 \}$ is not in $\mathscr{E}$. So I assume the contrary, then $B_r$ could be written as a finite union of $m$ disjoint intervals. $$ B_r = I_1 \cup \ldots \cup I_m $$ Then for a point $x \in Bball_r$ there exists $j \in \{ 1, \ldots, m \}$ such that $$ \sum_{i=1}^n x_i^2 < r^2 ~ \Leftrightarrow ~ ( a_1^{(j)} < x_1 < b_1^{(j)} ) \land \ldots \land ( a_n^{(j)} < x_n < b_n^{(j)} ) $$ where $a_i^{(j)}$ and $b_i^{(j)}$ denotes the bounds of the inteval $I_j$.

Okay, but how I proceed, I have a feeling that the upright formulae has a contradiction in it, but I am unable to make this formally?

EDIT1: Changed the definition of the ball from $\le$ (closed) to $<$ (open) (otherwise it would be trivial because a finite union of open sets is open and not closed)

EDIT2: Changed the naming from sphere to open ball (which is mathematical common usage for the set $\{x:|x| < r\}$.

EDIT3: Added the possiblity of $\le$ in the definition of an inteval, thanks to user Brusko's comment.

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There are some uncertainties about the problem. In formal mathematics, by a sphere with centre the origin one often means $\sum x_i^2=r^2$. For $\le r^2$, one usually writes ball. Also, you are using open intervals, but your sphere (or ball) is closed. That makes the problem too easy. Are you sure it is not the open ball that is meant? –  André Nicolas Nov 25 '12 at 21:36
    
yes, i changed it to be the open ball! –  Stefan Nov 25 '12 at 21:38
    
What about $S^1$? –  Isaac Solomon Nov 25 '12 at 21:43
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4 Answers

Check my answer to this question Prove that the unit open ball in $\mathbb{R}^2$ cannot be expressed as a countable disjoint union of open rectangles.. The same short argument can be used here

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Hey, thank. But i am sorry, i mean the open sphere! –  Stefan Nov 25 '12 at 21:36
    
I edited my answer to consider the open sphere –  Amr Nov 25 '12 at 21:40
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First the closure of both sides of $B_r = I_1 \cup \ldots \cup I_m$ to obtain the statement that the closed unit ball can be written as the finite union of closed intervals (closure of finite union is the finite union of closures). Now consider $(1,0,0..,0)$ - it's in the closed unit ball and thus in one of the intervals but in each of the intervals we can always change the second coordinate by some small number while still meeting the restraints whereas $(1,\epsilon,0,0,...,0)$ is not in the unit ball for $\epsilon\neq0$. (Of course, talking about the second coordinate requires $n$ to be at least 2, but the statement isn't true for $n=1$ anyway.)

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Let $n \geq 2$. If the open ball were the union of these finitely many open intervals, then the closed ball would be the union of the closure of these open intervals. Consider the points on the boundary of the closed ball. For a given open interval inside the sphere, how many possible boundary points can be in its closure? Can you show that there can only be finitely many? Do you see why this gives a contradiction?

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You claim that any union of a finite number of intervals could also be written as a union of a finite number of disjoint intervals. That doesn't seem correct to me. How about $$ A = \{ (x_1 , x_2) \in \mathbb{R}^2 , 0<x_1, x_2<2 \} \cup \{ (x_1 , x_2) \in \mathbb{R}^2 , 1<x_1, x_2<3 \} $$

Since by your definition all 'intervals' are open sets, and the set $A$ defined above is connected, but is not an interval, I see no way how it could possibly be expressed as a union of disjoint intervals.

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you are right, is added the possibility of $\le$ in the definition of an interval, then the statement holds. –  Stefan Nov 25 '12 at 22:08
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