How is this collection of open balls a covering of the first quadrant in the Euclidean $n$-space?

Question

Let $$ S \colon= \left\{ \left( x_1, \ldots, x_n \right) \in \mathbb{R}^n \colon \, x_1 > 0, \ldots, x_n > 0 \right\}. \tag{Definition 0} $$ For each positive real number $\alpha$, let $$ A_\alpha \colon= \left\{ \left( x_1, \ldots, x_n \right) \in \mathbb{R}^n \colon \sqrt{ \sum_{i=1}^n \left( x_i - \alpha \right)^2 } < \alpha \right\}, \tag{Definition 1} $$ that is, $A_\alpha$ is the open $n$-ball with center $( \alpha, \ldots, \alpha ) \in \mathbb{R}^n$ and radius $\alpha$. Finally, let $$ \mathscr{A} \colon= \left\{ A_\alpha \colon \alpha \in \mathbb{R}, \alpha > 0 \right\}. \tag{Definition 2} $$

Then how to show that the collection $\mathscr{A}$ covers $S$? That is, how to show that $$ S \subseteq \bigcup_{\alpha > 0} A_\alpha? \tag{0} $$ And, can we be even so precise as to state that $$ S = \bigcup_{\alpha > 0} A_\alpha? \tag{1} $$

My Attempt:

Let $\left( x_1, \ldots, x_n \right)$ be an arbitrary point of $S$. Then $x_1, \ldots, x_n \in \mathbb{R}$ such that $$ x_1 > 0, \ldots, x_n > 0. \tag{2} $$ Let us put $$ \alpha \colon= \frac{ x_1 + \cdots + x_n}{n}. \tag{Definition 3} $$ Then we note that \begin{align} \sqrt{ \sum_{i=1}^n \left( x_i - \alpha \right)^2 } &= \sqrt{ \left( \frac{(n-1) x_1 - x_2 \cdots - x_n}{n} \right)^2 + \left( \frac{-x_1 + (n-1)x_2 - \cdots - x_n}{n} \right)^2 + \cdots + \left( \frac{-x_1 - x_2 - \cdots - x_{n-1} + (n-1) x_n}{n} \right)^2 } \end{align}

Is what I have done so far correct? If so, then how to proceed from here?

Or, is there some other way around this problem?

Answer 1

This is false. You can visualize its falsity in the case $n=3$ by noticing that $A_\alpha$ is tangent to the $xy$ plane at the point $(\alpha,\alpha,0)$ which lies on the $45^\circ$ degree line on that plane; and similarly for the $yz$ and $zx$ planes. In particular, $A_\alpha$ never comes near points that lie on the $x$-axis, on the $y$-axis, or on the $z$-axis.

With that visualization as motivation, consider the case $n=3$ and the point $P = (1,0,0)$ which lies on the boundary of $S$. Let's find a positive constant $d>0$ such that each set $A_\alpha$ stays at least distance $d$ away from the point $P$.

Given $Q = (\alpha,\alpha,\alpha)$ with $\alpha>0$ we have $$d(P,Q) = \sqrt{3 \alpha^2 - 2\alpha + 1} $$ Let's compare $d(P,Q)$ to the radius $\alpha$ of the ball $A_\alpha$: $$d(P,Q) - \alpha = \sqrt{3 \alpha^2 - 2\alpha + 1} - \alpha \equiv f(\alpha) $$ The unique minimum value of the function $f(\alpha)$ occurs at $\alpha=2/3$ where that minimum is $f(2/3) = 1/3$. It follows that no matter what value of $\alpha>0$ we choose, the ball $A_\alpha$ never comes even within distance $d = 1/3$ of the point $P$. So for example the point $(1,.1,.1)$, whose distance from $P$ is $\sqrt{.02} < 1/3 = d$, is not contained in any ball $A_\alpha$.

Answer 2

Here's another perspective that may prove useful. The question of whether $\mathscr A$ covers $S$ is equivalent to asking whether for each $x \in S$, there exists an $\alpha > 0$ such that $\|x - \alpha e\| < \alpha$, where $\|\cdot\|$ denotes the Euclidean norm and $e$ denotes the all-ones vector $e = (1,1,\dots,1)$. This is in turn equivalent to asking whether for all $x \in S$, there exists an $\alpha > 0$ such that the function $$ f(\alpha) = \frac 1{\alpha}\|x - \alpha e\| = \left\|\frac 1{\alpha} x - e\right\| < 1. $$ Setting $t = 1/\alpha$, we can see that the above amounts to asking whether the distance from the open ray $R = \{tx : t > 0\}$ to the point $e$ is always strictly smaller than $1$.

For any $x \in S$, this distance can be readily computed with the help of a projection: we find that $$ d(e,R)^2 = \|e\|^2 - \frac{(e^Tx)^2}{x^Tx} = n - \frac{\left(\sum_{i=1}^n x_i\right)^2}{\sum_{i=1}^n x_i^2}. $$ By taking $x$ to be "sufficiently close" to one of the coordinate axes, the ratio ${\left(\sum_{i=1}^n x_i\right)^2}\Big /{\left(\sum_{i=1}^n x_i^2\right)}$ can be made arbitrarily close to $1$, so that $d(e,R)$ is made arbitrarily close to $n - 1$.

In the case of $n = 2$, this upper bound is equal to $1$, which means that $\mathscr A$ does cover $S$. For $n > 2$, that is no longer the case, which means that there exist points $x \in S \setminus \bigcup \mathscr A$.

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A potential followup question: what exactly does $U:= \bigcup \mathscr A$ look like, then? Because the membership of $x$ in $U$ depends only on the ray generated by $x$, we can conclude that $U$ is necessarily a linear cone. Moreover, it is easy to see that $U$ must be convex. So, the nature of $U$ can be determined by finding its intersection with the unit sphere $S^{n-1}$, and more specifically the boundary of this intersection within the sphere.

The boundary of the set $U \cap S^{n-1}$ consists of those points $x \in S^{n-1}$ satisfying the equation $$ 1 = n - \frac{\left(\sum_{i=1}^n x_i\right)^2}{\sum_{i=1}^n x_i^2} = n - \frac{\left(\sum_{i=1}^n x_i\right)^2}{1} \implies \sum_{i=1}^n x_i = n-1. $$ Thus, the boundary consists the intersection of $S^{(n-1)}$ with a plane, which is a circle whose axis is a line through the origin. With this, we can conclude that $U$ must be a right circular cone.

To find the angle of the cone, it suffices to find the angle between the cone's axis (namely $e$) and any point on the boundary. Since one point on the boundary of the cone is the point $b = (1,\dots,1,0)$, we find that $$ \cos \theta = \frac{b^Te}{\|b\|\,\|e\|} = \frac{n-1}{\sqrt{n(n-1)}} = \sqrt{\frac{n-1}{n}}. $$