Show that $S={\bigcup \limits_{n=1}^{\infty} \left(n - \frac{1}{n}, n + \frac{1}{n}\right)}$ cannot be a union of open balls of radius $r$

Question

I have to show that the set $$S={\bigcup \limits_{n=1}^{\infty} \left(n - \frac{1}{n}, n + \frac{1}{n}\right)}$$cannot be expressed as the union of open balls of the same radius $r$.

I am not sure how to prove this. I understand intuitively why this is not possible. But I am not sure how to proceed with the proof.

I know it's probably a proof by contradiction. But I am not sure how to begin - for example, I don't even know where to assume the balls are centred..

Answer 1

Notation: I will write $B(x,r)$ for the open ball of radius $r$ about $x$; i.e. $B(x,r)=(x-r,x+r)$. I will write $d(a,b)=|a-b|$ for the distance.

Suppose that $S$ is a union of balls of some fixed radius $r$; that is $S=\bigcup_x B(x,r)$ for some points $x\in S$. Certainly, there is some positive integer $N$ such that $r>\frac{2}{N}$. But then since $N\in S$ (as it lies in the interval $(N-\frac{1}{N},N+\frac{1}{N})$ ), $N\in B(x,r)$ for some $x$.

Note that on one hand, $x\in (N-\frac{1}{N}, N+\frac{1}{N})$. If not, then since $x\in S$, $x\in (n-\frac{1}{n}, n+\frac{1}{n})$ for some $n\neq N$. If $n>N$, then since $x>n-\frac{1}{n}>N$, $$r>d(x,N)\geq d(x,n-\frac{1}{n}),$$ we have $n-\frac{1}{n}\in B(x,r)$ yet $n-\frac{1}{n}\notin S$, which contradicts that $B(x,r)$ is a subset of $S$. If $n<N$, an entirely similar argument applies to get a contradiction from $n+\frac{1}{n}\in B(x,r)$, thus we must have $x\in (N-\frac{1}{N}, N+\frac{1}{N})$.

But then $$d(N\pm\frac{1}{N},x)\leq d(N\pm\frac{1}{N},N)+d(N,x)=\frac{1}{N}+d(N,x)<\frac{2}{N}<r,$$ so $N\pm \frac{1}{N}\in B(x,r)$ yet $N-\frac{1}{N}\notin S$, which contradicts that $B(x,r)$ is a subset of $S$.

Answer 2

Suppose we can write $$S=\bigcup_{i\in I}B_r(x_i)$$ for some radius $r$ and set of centers $\{x_i:i\in I\}$ with $I$ being an arbitrary index set. There exists $n\in\mathbb{N}$ such that $\frac 1 n<r$. Since $\left(n-\frac{1}{n}, n+\frac{1}{n}\right)\subset S$, we know $n\in S$. Notice also that $n-\frac 1 n\not\in S$ as $n-\frac 1 n>(n-1)+\frac{1}{n-1}$ and $n+\frac 1 n\not\in S$ as $n + \frac 1 n < (n+1)-\frac{1}{n+1}$. Since $n\in S$, there is some $i\in I$ such that $|x_i-n|<r$. We examine two cases:

In the first case, $x_i\le n$. Thus, we have $x_i-r<n-\frac 1 n<n<x_i+r$, so $n-\frac 1 n\in B_r(x_i)\subset S$, which is a contradiction.

In the second case, $x_i>n$. Thus, we have $x_i-r<n<n+\frac{1}{n}<x_i+r$. So $n+\frac 1 n\in B_r(x_i)\subset S$, which is also a contradiction.