This is a follow up question to the following question I asked. Sequencial chracterisation of $\limsup f(x)$ .
I would like to pove that
is equivalent to
Proof attempt
"$\Leftarrow$"
Let $\eta>0$. Consider $\limsup_{x\to p}f(x)-\frac{\eta}{2}$ and $\limsup_{x\to p}f(x)+\frac{\eta}{2}$. By thm. 1 (see below) we know that there is $\delta_{1}>0$ s.t. $f(x)<\limsup_{x\to p}f(x)+\frac{\eta}{2}$ and $\delta_{2}>0$ s.t. $f(x)>\limsup_{x\to p}f(x)-\frac{\eta}{2}$.
Define $\tilde{\delta}:=\min\{\delta_{1},\delta_{2}\}$ and consider $\epsilon<\tilde{\delta}$. Then we have that
$\begin{equation} \limsup_{x\to p}f(x)-\frac{\eta}{2}<f(x)<\limsup_{x\to p}f(x)+\frac{\eta}{2}\end{equation}$, $x\in E\cap B_{\epsilon}(a)\setminus\{a\}$, since $E\cap B_{\epsilon}(a)\setminus\{a\}\subset E\cap B_{\tilde{\delta}}(a)\setminus\{a\}$ and thus $|x|<\tilde{\delta}$.
Therefore $\limsup_{x\to p}f(x)+\frac{\eta}{2}$ is an upper bound for $f$ on $E\cap B_{\epsilon}(a)\setminus\{a\}$ and $\limsup_{x\to p}f(x)-\frac{\eta}{2}$ is a lower bound. Hence, we get the following chain of inequalities:
$$\begin{align}\limsup_{x\to p}f(x)-\frac{\eta}{2}&\leq\inf\{f(x)|x\in E\cap B_{\epsilon}(a)\setminus\{a\}\}\\ & \leq f(x)\leq\sup\{f(x)|x\in E\cap B_{\epsilon}(a)\setminus\{a\}\}\\ &\leq\limsup_{x\to p}f(x)+\frac{\eta}{2}\end{align}$$.
Hence,
\begin{align}\limsup_{x\to p}f(x)-\eta<\limsup_{x\to p}f(x)-\frac{\eta}{2}&\leq\sup\{f(x)|x\in E\cap B_{\epsilon}(a)\setminus\{a\}\}\\ &\leq\limsup_{x\to p}f(x)+\frac{\eta}{2}\\ &<\limsup_{x\to p}f(x)+\eta\\ &\iff |\sup\{f(x)|x\in E\cap B_{\epsilon}(a)\setminus\{a\}\}-\limsup_{x\to p}f(x)|<\eta\end{align}.
In other words we have that \begin{align}\forall\eta>0\exists\delta>0\forall\epsilon\in\mathbb{R}:|\epsilon|<\delta &\implies |\sup\{f(x)|x\in E\cap B_{\epsilon}(a)\setminus\{a\}\}-\limsup_{x\to p}f(x)|<\eta\\ &\iff \lim_{\epsilon\to 0}g(\epsilon)=\limsup_{x\to p}f(x)&,\\ & g(\epsilon):=\sup\{f(x)|x\in E\cap B_{\epsilon}(a)\setminus\{a\}\}\end{align}.
"$\implies$"
This direction is essentially done in "Introdunction to Mathematical Analysis" by Beatriz Laferriere, Gerado Laferriere,Nguyen Mau Nam.
Link: https://pdxscholar.library.pdx.edu/cgi/viewcontent.cgi?article=1003&context=pdxopen
There the following is stated:
i) From the first three lines of the Corollary I would conclude that $\limsup_{x\to p}f(x)\in A$. Thus, $A\neq\emptyset$
ii) The rest implies that $\limsup_{x\to p}f(x)$ is an upper bound of $A$. Therefore $\sup A$ exists.
Because $\limsup_{x\to p}f(x)\in A$ we have $\limsup_{x\to p}f(x)\leq\sup A$. But we also have that $\limsup_{x\to p}f(x)$ is an upper bound. Thus $\sup A\leq\limsup_{x\to p}f(x)$.Hence, $$\begin{align}\limsup_{x\to p}f(x)\leq\sup A\leq\limsup_{x\to p}f(x)\end{align}\iff\sup A=\limsup_{x\to p}f(x)$$
But since $\limsup_{x\to p}f(x)\in A$ we have that $\sup A=\max A$.
I hope it makes sense. It my be a bit overly detailed. Is there a shorter proof?
Many thanks in advacne!
Theorems used:
