I just wrote down the proof of the following easy proposition, and I was wondering about both the content (I would like to know if it is error-free), and the form of it.

Proposition: $\lim \inf A_n \subseteq \lim \sup A_n$.

Proof: Let $x \in \lim \inf A_n$ be arbitrary. Hence, there is a $m^* \in \mathbb{N}$ such that, for every $k \geq m^*$, $x \in A_k$. Proving that $x \in \lim \sup A_n$ is equivalent to state that for every $m \in \mathbb{N}$, there is a $j \geq m$ such that $x \in A_j$. Thus, let $m \in \mathbb{N}$ be arbitrary, and proceed by cases. If $m < m^*$, by defining $j:=m^*$ we have that $x \in A_{m^*}$. If $x \in A_m$, then it is enough to take define $j:=m+1$. Being the cases exhaustive, this establishes the result.

Any feedback is most welcome.
Thanks for your time.

  • 1
    $\begingroup$ This is correct. In the second case, $j=m$ would be ok. For a more compact version, without cases, use $j=\max(m^*,m)$. $\endgroup$ – Did May 9 '16 at 16:58
  • $\begingroup$ Thanks a lot for the feedback. Also, I actually posted the question because I had doubts concerning the cases. Indeed, I had another proof that ran at the end as "If $m < m^*$, by defining $j:=m^*$ we have that $x \in A_{m^*}$. If $m \geq m^*$, then the result follows from any choice of a natural number $i \geq m$. Hence, setting $j:=m$ implies that $x \in A_m$." This should be an extended version of your suggestion, right? $\endgroup$ – Kolmin May 9 '16 at 17:01

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