If I cover $\Bbb Q$ with open intervals, does it necessarily cover the whole $\Bbb R$ ? The answer is clearly no. The proof I was given says that since $\Bbb Q$ is countable, we can list its elements as a sequence $q_i$ and around each $q_i$ we take an open interval of length $\frac{\epsilon}{2^i}$. The total length will be equal to $\epsilon$.
But wasn't it enough to give the following counterexample : the covering $((2k-1) \pi,(2k+1) \pi), k \in \Bbb Z=\Bbb R \backslash({2\Bbb Z+1})\pi$ covers $\Bbb Q$ but not the whole $\Bbb R$.
Is there some even easier counterexample ?
Yes: $(-\infty,\pi)\cup(\pi,\infty)$, for instance.