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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?2012-11-25
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    yes, i changed it to be the open ball!2012-11-25
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    What about $S^1$?2012-11-25

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