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I have thought of a proof that the limit inferior of a sequence is less than or equal to the limit superior of that sequence as part of Exercise 6.4.3 part (c) from Tao's book Analysis I Fourth Edition. In particular, I approached this with a proof by contradiction, though I'm doubtful about it, and I attribute this to the fact that I've searched for other proofs of the same claim which, compared to what I came up with, are way more complicated. So that hints me that I've made a significant error, but I am somewhat convinced that I probably haven't, for the reason that I've exploited the entirety of part (a) (which already has a proof written down in the same book) which in and on itself is a property of liminf and limsup.

To begin, Tao denotes the liminf by $L^{-}$ and the limsup by $L^{+}$, and so it is asked by the reader to show that $L^{-} \le L^{+}$ among other two inequalities which were quite trivial to prove and so I won't bother mentioning them.

In addition, part (a) states that for a given sequence of real numbers with a starting index m denoted by $(a_n)_{n=m}^{\infty}$ we have that:(1) For every $x\gt L^{+}$, there exists an $N\ge m$ such that $a_n\lt x$ for all $n\ge N$. Similarly, (2) for every $y\lt L^{-}$, there exists an $N\ge m$ such that $a_n\gt y$ for all $n\ge N$.

Proof (of part (c)). Assume for the sake of contradiction that $L^{-}\gt L^{+}$. Thus by (1) we can find some $n_0\ge m$ such that $a_n\lt L^{-}$ for all $n\ge n_0$. Observe that $a_{n_0+k}\lt L^{-}$ for all non-negative integers $k\ge 0$. Hence, by (2) we can find another $n_0^{'}\ge m$ such that $a_n\gt a_{n_0+k}$ for all $n\ge n_0^{'}$. Now by the trichotomy of the natural numbers, we have the cases that $n_0\ge n_0^{'}$ or $n_0\lt n_0^{'}$. Now, if $n_0\ge n_0^{'}$, then $a_{n_0}\gt a_{n_0+k}$ for all $k\ge 0$. Thus, by setting $k=0$ we obtain $a_{n_0}\gt a_{n_0}$, a contradiction. On the other hand, if $n_0\lt n_0^{'}$, then we can find some natural number $\ell$ such that $n_0+\ell\gt n_0^{'}$. Therefore, $a_{n_0+\ell}\gt a_{n_0+k}$ for all $k\ge 0$. Thus, by setting $k=\ell$ we obtain $a_{n_0+\ell}\gt a_{n_0+\ell}$, a contradiction. So we see that in every case we encounter a contradiction which means that the original assumption that $L^{-}\gt L^{+}$ is utterly false, and so we must have that $L^{-}\le L^{+}$, q.e.d.

So I ask, is the above proof valid or total gibberish?

Vaskara_GRek_O
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    Use the sequences $a^+n := \sup{k>n} a_k$ and $a^-n := \inf{k>n} a_k$, you have $a^+_n \geq a^-_n$ for all $n$... this will be much clearer and easier since $L^+ = \lim a^+_n = \inf a^+_n$ and $L^- = \lim a^-_n = \sup a^-_n$. – julio_es_sui_glace Jul 03 '23 at 18:15
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    Honestly, a bit hard to follow, and I believe unnecessarily complicated. Why not just say that (assuming as you do $L^+<L^-$) if you call $A$ the midpoint of $L^+$ and $L^-$, all the elements of the sequence have to be both greater and smaller than $A$ after some point (using the same properties you called (1) and (2)) which gives a contradiction right away? – GReyes Jul 03 '23 at 18:18
  • @GReyes You are right, the contradiction you are proposing is much more immediate. I didn't think it could be simpler. Although, I am still looking for that confirmation. Besides that, you've also mentioned that it was difficult to follow the proof, any tips as to how I could make my future proofs more comprehensible and aesthetically pleasing, for the sake of my mathematical training? I'd really appreciate it. – Vaskara_GRek_O Jul 03 '23 at 21:28
  • I read your proof in detail and I think it is correct. I don’t have an answer for your question on how to produce more pleasing proofs. For me it is a matter of personal preference. I am a visual thinker. If I don’t “see” the proof all at once, I am not satisfied. In other words, a long chain of otherwise correct steps may be a good proof for some people, but I am not quite satisfied if I am not able to extract the main idea, which can be expressed in a few lines. – GReyes Jul 05 '23 at 16:44

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