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In my homework it is asked to prove the following statement:

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

Is this equivalent to proving:

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

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1 Answer 1

The statements are absolutely not equivalent. On the one hand, it's clear by definition of $y$ as a limit point that there's some sequence in $X \setminus \{y\}$ which tends to $y$. On the other hand, it's not clear that this sequence will tend to the lim sup of f as $x \to y$, or even that any sequence satisfies this property.

To proceed, first assume that $\limsup_{x \to y} f(x) < \infty$, and call this limit $A$. For each positive integer $k$, find a $\frac{1}{k} > \delta_k > 0$ so that $|A - \sup\{f(x): x\in X \setminus \{y\} \cap B(y,\delta) \}| < \frac{1}{k}$, for $0 < \delta < \delta_k$. So pick some $x_k$ (and we can arrange the situation so that the $x_k$ are distinct, though it's not strictly necessary) in $X \cap \{y\}$ with $|x - y| < \frac{1}{k}$ so that $|A - f(x_k)| < \frac{1}{k}$.

Then we must have that $x_k \to y$ since each sits inside of a ball $B(y, \frac{1}{k})$. And it's also clear by construction that $f(x_k) \to A$ as $k \to \infty$.

This outline can be easily modified to when $A = \infty$.

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