Prove the existence of such sequence

Question

Let $X \subseteq \mathbb{R}^m$, let $y$ be a limit point of $X$, and let $f:X\setminus\{y\} \to \mathbb{R}$ be a function.

I need to prove that:

There is a sequence $(x_k)$ of points from $X \setminus \{y\}$ such that $\lim _{k \to \infty}x_k=y$ and $\lim_{k\to \infty}f(x_k)=\overline \lim_{x\to y}f(x)$.

Here $\displaystyle \overline \lim_{x\to y}f(x) = \lim_{\delta \to 0^+} \sup\{f(x): x \in X\setminus \{y\} \cap B(y,\delta)\}$.

I got confused since the sequence need to satisfy two conditions and how I can construct my proof. Any hints, please.

Answer 1

I am assuming $f$ is bounded above in $I - \{y\}$ for some non-empty interval $I$ containing $y$ in its interior.

For $ \delta > 0$ let $g(\delta) = \sup_{ \{x: 0 < | x - y | < \delta\} } f(x) $. Then $\inf_{ \delta > 0} g(\delta) = \lim_{ \delta \to 0^+ } g(\delta) = \limsup_{x\to y} f(x).$ Note $g$ is a increasing function hence for any $(\delta_n)$ which strictly decreases to $0$ $ \lim_{n \to \infty} g(\delta_n ) = \limsup_{x\to y} f(x) $, Next note for any $g( \delta_n ) < \infty $ for sufficiently large $n$ and for such $n$ there exists a $x_{n}$ with $ 0 < |x_n - y| < \delta_n $ such that $ g(\delta_n) - \frac{1}{n} \leq f(x_n) \leq g(\delta_n) $ , this follows from well known properties of the supremum.

If $f$ is not bounded above in every $I-\{y\}$ then $g(\delta) = \infty$ for every $\delta > 0$ and we can find a sequence $x_n$ with $f(x_n) > n$ and $ 0 < |y - x_n | < 1/n $. (This post was updated to plug a hole in the previous proof noticed by proximal).