Does it seem plausible to conjecture that given an enumeration of the rational numbers $\{ q_1, q_2, q_3, \ldots \}$ and a convergent sequence $\{ \epsilon_n \}_{n \in \mathbb{Z^+}} \subseteq \mathbb{R^+}$ such that $\lim \epsilon_n = 0$ but $\sum_{n \in \mathbb{Z^+}} \epsilon_n = +\infty$, the set of open intervals $\{ (q_n - \epsilon_n, q_n + \epsilon_n) \mid n \in \mathbb{Z^+} \}$ covers $\mathbb{R}$? For instance, can $\mathbb{R}$ be covered by $\{ (q_n - \frac{1}{n}, q_n + \frac{1}{n}) \mid n \in \mathbb{Z^+} \}$? It looks like if we allow $\lim \epsilon_n \neq 0$, the conjecture might work...
Asked
Active
Viewed 76 times
1 Answers
1
I don't think so.
Separate the index set $\mathbb Z^+$ into two parts: $N_1 = \{ n \in \mathbb Z^+ : n = 2^k\text{ for some } k \in \mathbb Z^+\}$ and $N_2 = \mathbb Z^+ \setminus N_1$.
Enumerate the rational numbers as follows: if $n \in N_2$ let $q_n = n$, and let $\{q_n\}_{n \in N_1}$ be an arbitrary enumeration of $\mathbb Q \setminus N_2$. Let $\epsilon_n = \frac 1n$ for all $n$.
The intervals $(n-\frac 1 n,n+ \frac 1n)$, $n \in N_2$, fail to cover a portion of the line containing intervals of infinite total length.
On the other hand, the intervals $(q_n - \frac 1n, q_n + \frac 1n)$, $n \in N_1$, may be indexed as $(q_{2^k}-\frac 1{2^k},q_{2^k}+ \frac 1{2^k})$, $k \in \mathbb Z^+$, and thus have finite total length. This means they can't cover everything the first set of intervals missed.
Umberto P.
- 52,165
-
Perfect. I think that ${ q_n }_{n \in N_1}$ is an arbitrary enumeration of $\mathbb{Q} \setminus N_2$. – brandao Apr 05 '18 at 15:23
-
Quite right. I'll fix it. – Umberto P. Apr 05 '18 at 16:01