There is a theorem in Ross' book Elementary Analysis involving limsup and lim inf. The portion of the theorem that im asking about states: If liminf$s_n$=limsup$s_n$, then lim$s_n$ is defined and lim$s_n$=liminf$s_n$=limsup$s_n$.

The portion of the proof that I'm confused about proceeds as follows: Suppose that liminf$s_n=s$ where $s$ is a real number. We need to prove lim$s_n=n$. Let $\epsilon$>0. Since s=lim$v_N$ (where $v_N=sup\{s_n:n>N\}$) there exists a positive interger $N_0$ such that

$|s-sup\{s_n:n>N_0\}|$<$\epsilon$. Thus $sup\{s_n:n>N_0\}<s+\epsilon$, so

$s_n<s+\epsilon$ for all $n>N_0$. 1)

Similarly, there exsists $N_1$ such that $|s-inf\{s_n:n>N_1\}<\epsilon$, hence $inf\{s_n:n>N_1\}>s-\epsilon$, hence

$s_n>s-\epsilon$ for all $n>N_1$ 2).

From 1) and 2) we conclude $s-\epsilon<s_n<s+\epsilon$ for $n>max\{N_0,N_1\}$ equivalently $|s_n-s|<\epsilon$ for all n>$max\{N_0,N_1\}$. This proves $lims_n=s$ as desired.

My question is about the first portion of the proof, right after he introduces $N_0$, how does $|s-sup\{s_n:n>N_0\}$<$\epsilon$ imply $sup\{s_n:n>N_0\}<s+\epsilon$?

  • $\begingroup$ sorry, I forgot to add the other line to complete the absolute value. $\endgroup$ – Oscar Flores Mar 19 '14 at 15:11

So $$|s-\sup\{s_n:n>N_0\}|<\epsilon$$ can be written as $$-\epsilon -s<-\sup\{s_n:n>N_0\}<\epsilon-s$$ or equivalently $$s-\epsilon<\sup\{s_n:n>N_0\}<s+\epsilon$$ which gives you the upper bound that the author uses in his subsequent calculations.


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