# For $s_n$ a sequence in $\Bbb R$, if $\lim s_n$ defined as a real number, then $\liminf s_n = \lim s_n = \limsup s_n$.

I'm studying some real analysis, and I'm trying to figure out how to prove the following theorem. For the most part, I've got things figured out. I'll post the theorem and my proof (which closely "follows" that of Kenneth Ross's proof within Elementary Analysis: The Theory of Calculus, Second Edition).

Let $(s_n)$ be a sequence in $\Bbb R$. If $\lim s_n$ is defined as a real number, then $\liminf s_n$ = $\lim s_n$ = $\limsup s_n$.

After dealing with other cases, the proof proceeds as follows:

Now suppose $\lim s_n = s$, where $s$ is a real number. Consider $\epsilon > 0$. There exists a positive integer $N$ such that $|s_n - s| < \epsilon$ for $n > N$. Thus $s_n < s + \epsilon$ for $n > N$, so $v_n = \sup\{s_n : n> N\} \le s + \epsilon$. Also, $m > N$ implies $v_m \le s + \epsilon$, so $\limsup s_n = \lim v_m \le s + \epsilon$. Since $\limsup s_n \le s = \lim s_n$ for all $\epsilon > 0$, no matter how small, we conclude $\limsup s_n \le \liminf s_n$.

Now, things go awry right about here:

A similar argument shows $\lim s_n \le \liminf s_n$.

I understand the proof up until this point. Basically, our sequence $(s_n)$ hits a "capstone" since $\lim s_n = s$ (eventually, considering $n > N$ in the limit definition and our other inequalities). Therefore our $\limsup s_n$ should either hit less than the capstone or the capstone, never higher.

However, proving that $\lim s_n \le \liminf s_n$ seems impossible to me, especially using a similar argument. I can't even follow through with a proper proof. I'm stuck on how this is even possible. I know why they want to set this equality up (it allows us to achieve $\liminf s_n = \lim s_n = \limsup s_n$ since $\liminf \le \limsup s_n$). But how they get there is beyond me. Using a similar argument, I arrive here:

As before, since $|s_n - s| < \epsilon$, $s_n < s + \epsilon$ for $n > N$. Let $u_N = \inf\{s_n: n > N\}$ $\le s + \epsilon$.

Which seems useless given that $\liminf s_n \le \limsup s_n$. Wouldn't that imply $\liminf s_n \le \lim s_n = s$? Furthermore, since the infimum is defined as the greatest lower bound of a set, and the $\liminf s_n$ can only ever possibly reach a limit of $s$, how can the statement

$\lim s_n \le \liminf s_n$

even be true? I sense that I'm missing something obvious here, but I've been bugged about this argument for a while...

• Assume $\lim(s_n)>\liminf(s_n)$ and derive a contradiction. It follows almost immediately from the definition of limit and liminf. – Matthew Levy Nov 25 '14 at 0:33
• Alright, here goes: $lim s_n > \liminf s_n$. As $\lim_{N\to \infty}$, $s > s$, thus revealing the contradiction? – 6nop Nov 25 '14 at 0:45
• Ah, so then it follows that $\liminf_{N \to \infty} s_n \ge \lim_{N \to \infty} s_n$, which isn't a contradiction since $s \ge s$. Thanks! – 6nop Nov 25 '14 at 16:19

$|s_n-s|<\epsilon$ implies $s_n<s+\epsilon$ and $s_n>s-\epsilon$ for all $n>N$. So $\inf\{s_n: n>N\}$ is bounded below by $s-\epsilon$. Now take $\epsilon$ to zero...