Find the $\limsup\limits_{m \rightarrow \infty} [0,\frac{1}{m}]$

Question

The question is to find the following: \begin{equation} \limsup\limits_{m \rightarrow \infty} \big[0,\frac{1}{m}\big] \end{equation} What I have started with is using the result of the Borel-Cantelli Lemma, i.e. if $\{E_m\}$ is a set of Lebesgue measurable sets in $\mathbb{R}^n$, then: \begin{equation} \limsup\limits_{m \rightarrow \infty}E_m = \bigcap^{\infty}_{q=1}\bigcup^{\infty}_{m=q}E_m \end{equation} From here, can we just say that, by using this lemma, the answer is $\{0\}$. Not sure how to formalise this idea, any help or thoughts are appreciated.

Answer 1

Let $m \in \mathbb{N}$; let $x \in \cup_{n \geq m}[0, 1/n]$. Then $x \in [0,1/n]$ for some $n \geq m$. Note that $n \geq m$ iff $\frac{1}{n} \leq \frac{1}{m}$; so we have $x \in \{ 0 \}$.

Answer 2

The key of using this lemma (I think it's a definition actually) is to realizing which elements are not in the limit superior.

You can see that negative numbers aren't in any of the sets, so aren't in the limit superior either.

Given a real number $r > 0$, you can see that there is a number $M$ large enough such that $r > \frac 1M$. Then, $r \notin \left[0,\frac 1m\right]$ for any $m \geq M$.

This means, therefore, that $r \notin \bigcup_{m = M}^\infty \left[ 0,\frac 1m \right]$, which means that $r \notin \bigcap_{q=1}^\infty \bigcup_{m=q}^\infty \left[ 0,\frac 1m\right]$, so that $r \notin \limsup \left[ 0,\frac1m \right]$.

Hence, only $r=0$ remains. I leave you to see that since $r \in \left[ 0,\frac 1m\right]$ for all $m$ , it is in the limit superior. Use the definition, the way I did.