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Theorem 1.3.3: Let $L \in R$ and let $x_n$ be a sequence of real numbers. Then $x_n \to L$ if and only if $\limsup x_n = \liminf x_n = L$.

Proof: First, suppose $x_n \rightarrow L.$ Thus if $\epsilon > 0$ there exists $N \in \mathbb{N}$ such that $ n \geq N$ implies $|x_n-L| < \epsilon/2$, which implies $s_n = \sup(T_n) \leq L + \epsilon/2$ and $i_n = \inf(T_n) \geq L-\epsilon/2$.

Thus

$$L-\frac{\epsilon}{2} \leq i_n \leq s_n \leq L +\frac{\epsilon}{2}$$

which implies that $|s_n-L| \leq \frac{\epsilon}{2} \lt \epsilon$ and $|i_n-L| \leq \frac{\epsilon}{2}<\epsilon$, for all $\epsilon > 0$. Thus $s_n \rightarrow L $ = lim sup $x_n$ and $i_n \rightarrow L$ = lim inf $x_n$.

I only included the first part of the proof. I boxed the part where I was confused. Why did they use inequalities

$s_n = \sup(T_n) \leq L + \epsilon/2$ and $i_n = \inf(T_n) \geq L-\epsilon/2$ instead of

$s_n = \sup(T_n) \lt L + \epsilon/2$ and $i_n = \inf(T_n) \gt L-\epsilon/2$.

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  • $\begingroup$ Do strict inequalities matter? If the strict inequality is true, then the given inequality is also true. $\endgroup$
    – Michael Burr
    Commented Dec 29, 2016 at 20:02
  • $\begingroup$ @HenryW. These are just upper and lower bounds. As written, one could just vary $N$ or $\varepsilon$ to turn a strict inequality into a non-strict inequality. $\endgroup$
    – Michael Burr
    Commented Dec 29, 2016 at 20:04
  • $\begingroup$ @MichaelBurr I can see that the proof will still work if we used either strict or non-strict inequalities, I just wanted to make sure I know the reason behind it. Could it be because of the fact that least upper bound can be equal to the upper bound? $\endgroup$
    – user3000482
    Commented Dec 29, 2016 at 21:06
  • $\begingroup$ It's hard to say without $T_n$ being defined, but it may just be that the author of the proof is being lazy. $\endgroup$
    – Michael Burr
    Commented Dec 29, 2016 at 22:49

2 Answers 2

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Maybe you are thinking that the strict inequality is valid because $|x_n-L|<\epsilon/2 $ is true, but note that this inequality is true for any element $x_n$ in the sequence beyond $N$, but the supremum or infimum of such elements in $T_n$ may not be any of the elements in the tail of the sequence, so you need to write it as a weak inequality. Also note that this inequality does not affect the proof anyway.

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  • $\begingroup$ Yes, you are right. I thought they should have used the strict inequality because we chose $N$ such that for $n \geq N$, $x_n$ would be within the $L + \frac{\epsilon}{2}$ and $L - \frac{\epsilon}{2}$. But since $s_n$ and $i_n$ are one of $x_n$ for some $n \geq N$, by our assumption, shouldn't the inequality for $i_n$ and $s_n$ be also strict as for $x_n$? $\endgroup$
    – user3000482
    Commented Dec 29, 2016 at 21:02
  • $\begingroup$ "But since $s_n$ and $i_n$ are one of the $x_n$ for some $n\geqN$, by our assumption" -- There is no assumption in the statement that guaranties this. $\endgroup$
    – Regio
    Commented Dec 30, 2016 at 15:31
  • $\begingroup$ The assumption that $x_n \rightarrow L$? $\endgroup$
    – user3000482
    Commented Dec 30, 2016 at 18:28
  • $\begingroup$ Nop, the fact that $x_n\rightarrow L$ does not guarantee that any element in the sequence is equal to the lim, the lim sup or the lim inf. $\endgroup$
    – Regio
    Commented Jan 2, 2017 at 21:53
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There is a reason to not use strict inequalities. For arbitrary $S \subseteq \mathbb{R}$, it could happen that $x < L$ for all $x \in S$, but $\sup S = L$. (but never $\sup S > L$)

For example, take $S = \left]0,1\right[$ and $L = 1$.

However, since $x \to L$ is assumed, you can safely replace $\leq$ with $<$ since the limiting case $\sup T_n = L + \epsilon/2$ never happens.

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