I am working through Elementary Analysis by Ross and don't understand part of his proof of Theorem 10.7 (i). This question is in a similar vein as this one, but regarding a different part of the proof.
Theorem 10.7
Let ($s_n$) be a sequence in $\mathbb{R}$.
(i) If lim $s_n$ is defined [as a real number, +∞, or −∞], then lim inf $s_n$ = lim $s_n$ = lim sup $s_n$.
Proof
We use the notation $u_N$ = inf {$s_n$:n>N}, $v_N$ =sup {$s_n$:n>N}, u = lim $u_N$ = lim inf $s_n$ and v = lim $v_N$ = lim sup $s_n$.
Suppose lim $s_n$ = +∞. Let M be a positive real number. Then there is a positive integer $N$ so that $n>N$ implies $s_n>M$.
Then $u_N$ = inf {$s_n:n>N$} $≥ M$. It follows that $m>N$ implies $u_m ≥ M$. In other words, the sequence ($u_N$) satisfies the condition defining lim $u_N = +∞$, i.e., lim inf $s_n = +∞$. Likewise lim sup $s_n = +∞$.
The case of lim $s_n = −∞$ is handled in a similar manner.
Now suppose lim $s_n = s$ where $s$ is a real number. Consider $ϵ > 0$. There exists a positive integer N such that |$s_n − s$| $< ϵ$ for $n > N$. Thus $s_n < s + ϵ$ for $ n > N$, so
$v_N$ = sup {$s_n : n > N$ } $≤ s + ϵ$ .
Also, $m > N$ implies $v_m ≤ s + ϵ$, so lim sup $s_ n$ = lim $v_m ≤ s + ϵ$. Since lim sup $s_n ≤ s + ϵ$ for all $ϵ > 0$, no matter how small, we conclude lim sup $s_n ≤ $lim $s_n$. A similar argument shows lim $s_n ≤ $lim inf$s_n$. Since lim inf $s_n ≤$ lim sup$s_n$, we infer all three numbers are equal:
lim inf $s_n$ = lim $s_n$ = lim sup $s_n$.
My question is in regards to this statement towards the end:
"Since lim sup $s_n ≤ s + ϵ$ for all $ϵ > 0$, no matter how small, we conclude lim sup $s_n ≤ $lim $s_n$."
Why may we make this assumption? For example, let lim sup $s_n = s + ϵ$. Then since $ϵ > 0$, no matter how small, shouldn't we conclude lim sup $s_n > $ lim $s_n$ in that case?
Thank you!
