TITLE: Generalizing the notion of limit of a real valued function to the $\limsup$
QUESTION [0 upvotes]: We have the following proposition in Real Analysis

Let $X\subseteq\mathbb{R}$, let $f: X\rightarrow\mathbb{R}$ be a function, let $E\subseteq X$, let $x_{0}\in\overline{E}$ and let $L\in\mathbb{R}$. Then the following are equivalent: $(a)$ $\lim_{x\rightarrow x_{0};x\in E} f(x) = L$. $(b)$ For every sequence $(a_{n})_{n = 0}^{\infty}$ of elements in $E$ such that $\lim_{n\rightarrow\infty}a_{n} = x_{0}$ we have $\lim_{n\rightarrow\infty}f(a_{n}) = L$.

The limit superior of a real-valued function is defined as $\limsup_{x\rightarrow x_{0};x\in E}f(x) = \inf_{\varepsilon >0}\sup_{|x-x_{0}| < \varepsilon; x\in E}f(x)$. Is the following generalization of the above proposition true?

Let $X\subseteq\mathbb{R}$, let $f: X\rightarrow\mathbb{R}$ be a function, let $E\subseteq X$, let $x_{0}\in\overline{E}$ and let $L\in\mathbb{R}$. Then the following are equivalent: $(a)$ $\limsup_{x\rightarrow x_{0};x\in E} f(x) = L$. $(b)$ For every sequence $(a_{n})_{n = 0}^{\infty}$ of elements in $E$ such that $\lim_{n\rightarrow\infty}a_{n} = x_{0}$ we have $\limsup_{n\rightarrow\infty}f(a_{n}) = L$.

Thoughts: This is what I have got so far: Suppose $(a)$ holds and let $(a_{n})_{n = 0}^{\infty}$ be a sequence of elements in $E$ such that $\lim_{n\rightarrow\infty}a_{n} = x_{0}$. Then for every $\varepsilon > 0$ there exists $N_{\varepsilon}\in\mathbb{N}$ such that for all $n\geq N_{\varepsilon}$ we have $a_{n}\in\{x\in E: |x-x_{0}| < \varepsilon\}$. Therefore, for all $n\geq N_{\varepsilon}$ we have $f(a_{n})\leq \sup\{f(x)\in E: |x-x_{0}| < \varepsilon\}$.
$$\implies \limsup_{n\rightarrow\infty}f(a_{n})\leq \sup_{|x-x_{0}| < \varepsilon; x\in E}f(x)$$
Since this inequality is true for every $\varepsilon > 0$ we get
$$\limsup_{n\rightarrow\infty}f(a_{n})\leq \inf_{\varepsilon > 0}\sup_{|x-x_{0}| < \varepsilon; x\in E}f(x) = \limsup_{x\rightarrow x_{0};x\in E}f(x) = L$$
If the proposition is true, how would one go about showing the reverse inequality and also the converse (i.e. $(b)\implies (a)$)?

REPLY [1 votes]: Here is a partial answer based on the comment by @copper.hat. One can show that if $\limsup_{x\rightarrow x_{0};x\in E}f(x) = L$ then for every sequence of elements $(a_{n})_{n = 0}^{\infty}$ in $E$ that converges to $x_{0}$ we have $\limsup_{n\rightarrow\infty}f(a_{n})\leq L$ and there exists a sequence such that $\lim_{n\rightarrow\infty}f(a_{n}) = L$.
$L = \inf_{\varepsilon > 0}\sup_{|x-x_{0}| < \varepsilon; x\in E}f(x)$. Suppose there exists $\varepsilon > 0$ such that $\sup_{|x-x_{0}| < \varepsilon; x\in E}f(x) = L$. Then for every $\varepsilon'\leq\varepsilon$ we have $\sup_{|x-x_{0}| < \varepsilon'; x\in E}f(x) = L$. Then one may construct a sequence $(\varepsilon_{n})_{n = 1}^{\infty}$ going to $0$ such that $\varepsilon_{n}\leq \varepsilon$, and a sequence $(x_{n})_{n = 1}^{\infty}$ such that $x_{n}\in\{x\in E:|x -x_{0}| < \varepsilon_{n}\}$ and $L-1/n< f(x_{n})\leq L$ (by the axiom of choice) for each $n\in\mathbb{N}\setminus\{0\}$. Then $\lim_{n\rightarrow\infty}x_{n} = x_{0}$ and $\lim_{n\rightarrow\infty}f(x_{n}) = L$.
On the other hand, if there is no $\varepsilon > 0$ such that the infimum is attained then for each $n\in\mathbb{N}\setminus\{0\}$ we may pick $\varepsilon_{n}> 0$ (by the axiom of choice) such that $L < \sup_{|x-x_{0}| < \varepsilon_{n}; x\in E}f(x) < L+1/n$. We claim that $\lim_{n\rightarrow\infty}\varepsilon_{n} = 0$. Suppose this is false for the sake of contradiction, then there exists $\varepsilon > 0$ such that for every $N\in\mathbb{N}\setminus\{0\}$ there exists $n\geq N$ such that $\varepsilon_{n} > \varepsilon$. For such $n$ we would then have $L+1/n>\sup_{|x-x_{0}|<\varepsilon; x\in E}f(x)$. However, by the Archimedian property  there exists $M\in\mathbb{N}\setminus\{0\}$ such that $M\big(\sup_{|x-x_{0}|<\varepsilon; x\in E}f(x)-L\big) > 1$ which contradicts the previous statement. Then one may pick $(x_{n})_{n = 0}^{\infty}$ as before such that $L < f(x_{n})<L+1/n$ which is the desired sequence.
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