Showing a biconditional statement about function lim sups in $\Bbb R^n$, and codifying the intuition into a proof $
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$The Statement: Let $f : E \subseteq \R^n \to \R$. The claim to prove:
$$\beta = \limsup_{x \to x_0} f(x) \iff \text{conditions (i) & (ii) below hold}$$
The conditions:

*

*(i) $\exists \set{x_n}_{n \in \N} \subseteq \R^n$ a sequence such that $f(x_n) \to \beta$

*(ii) $(\forall b > \beta)(\exists \delta > 0)(f(x) < b \text{ for } x \in B^* \cap E)$

Notable Definitions & Notations: I think the more relevant definitions to go over would be these two:

*

*We denote the "punctured ball" centered at $x_0$ with radius $\delta$ by $B^*$, and more explicitly may also write
$$B^* := B^*(x_0;\delta) := B(x_0;\delta) \setminus \set{x_0}$$
(Some of you may be more familiar with the notation $B_\delta(x_0)$ to denote a ball of radius $\delta$ centered at $x_0$.) Essentially the punctured ball $B^*$ is the usual ball with the center removed.


*We define the limit supremum of $f$ as $x \to x_0$ as so:
$$\limsup_{x \to x_0} f(x) = \lim_{\delta \to 0} \left( \sup_{x \in B^* \cap E} f(x) \right)$$

Context & Attempts: Ultimately this is a homework problem, so I would prefer to not have full proofs given out; moreso just nudges and such.
Now, in the senses of geometry, visuals, and intuition, I think I have some ideas as to what is going on, though not quite the whole picture.

*

*In the forward direction: let $\beta$ be $\limsup f(x)$. Then that means two things. Firstly, we can find a sequence $x_n$ in the domain where that sequence under $f$, $f(x_n)$, converges to $\beta$. I think that helps get $\beta \le \sup f(x)$ on the relevant set. Moreover, whenever we have a number $b$ larger than $\beta$, then condition (ii) essentially constricts $b$ and $\beta$ in a sense. The closer $b$ and $\beta$, the smaller $\delta$ is, and eventually I think that will help us get to equality.


*In the backwards direction: we know there is a sequence $\set{x_n}$ where $f(x_n) \to \beta$, and whenever $b > \beta$, $x \in B^*(x_0;\delta) \cap E$ have $f(x) < b$ (though where $\beta$ lies between the two is not known). I think this would give that $\sup f(x) \le b$ for the relevant set, and as $\delta \to 0$, $b$ and $\beta$ get closer and closer together until we somehow get equality.
My issues with these are more like ... trying to codify my intuitions. I have written down some things, but I'm not too certain they're on the right track.
For instance, for the forward direction:

Attempt at the Forward Direction:
Suppose $f : E \subseteq \R^n \to \R$ and $\beta := \displaystyle \limsup_{x \to x_0} f(x)$. Then by definition,
$$
\beta = \lim_{\delta \to 0} \para{ \sup_{x \in B^*(x_0;\delta) \cap E} f(x) }
$$
Let $\delta_k := 1/k$. Then define $\BB_k := B^*(x_0;\delta_k)$. Since $\delta_k \to 0$, it holds that
$$
\sup_{x \in \BB_k \cap E} f(x) \to \beta
$$
Thus, define $x_k$ by $x_k \in \BB_k \cap E \subseteq E$, and then $\set{x_k}_{k \in \N}$ is one such that $f(x_k) \to \beta$, satisfying condition (i).
Suppose $b > \beta$. To see condition (ii) holds, suppose otherwise, that $\forall \delta > 0$, $f(x) \ge b$ for $x \in B^* \cap E$. This would mean that
$$
\sup_{x \in B^*(x_0;\delta) \cap E} f(x) \ge b > \beta
$$
for all $\delta > 0$ and in particular the case $\delta \to 0$. This would give that $ \displaystyle\limsup_{x \to x_0} f(x) = b$ instead from the definition, contradicting that $\beta$ is the lim sup. Thus, a contradiction is reached and condition (ii) holds.

Attempt at the Backwards Direction: (which is moreso scratch work that stalled hard)
We're going to suppose (i) & (ii) hold. Since $\exists \set{x_k}_{k \in \N} \subseteq E$ such that $f(x_k) \to \beta$, from definition, $\forall \ve > 0$, $\exists N \in \N$ such that, $\forall n \ge N$, $|f(x_n) - \beta| < \ve$. In particular, we can consider the elements of $\set{x_k}$ which are within a radius $\delta > 0$ of $x_0$ and are different from $x_0$; these elements define a subsequence $\set{x_{n_k}}_{k \in \N} \subseteq B^* \cap E$.
From the fact that limits are unique, and if a sequence $z_k \to L$, then all subsequences of $\set{z_k}_{k \in \N}$ converge to $L$ as well. Thus, $\set{x_{n_k}}_{k\in\N}$ is a sequence in $B^* \cap E$. If we take the limit as $\ve,\delta \to 0$, it then follows from these and the definition of supremum that
$$
\lim_{\delta \to 0} \para{ \sup_{B^* \cap E} f(x) } \le \beta
$$
Consider condition (ii) and that for a $b > \beta$, $\exists \delta > 0$ such that $f(x) < b$ for $x \in B^* \cap E$.
(That's where I get lost though...)

Questions & Concerns:

*

*Is my intuition for what's going on correct?

*If not, what's the correct intuition and the general approach I should use?

*Is there anything remotely salvageable from my approaches?

*If there are errors with my approach, what are they? How might I rectify them?

Thanks for any insight you can give me!

Edit: (thoughts for converse as of 1/27/2021)
I had a thought for the approach that seems to neatly tie conditions (i) and (ii) together.
Let $\set{b_k}_{k \in \N}$ be a monotone decreasing sequence with limit $\beta$, i.e. $b_k > \beta$ for every $k$, and $b_k \searrow \beta$. Define $\ve_k := b_k - \beta$.
Consider the preimage $f^{-1}(\beta,\ve_k)$. This may consist of a set of disjoint sets, so consider only the one $\mathcal{A_k}$ the point $x_0$ lies in. For each preimage determined by $\ve_k$, we can get a $\mathcal{A_k}$, containing a ball $B(x_0;\delta_k)$.
Thus, each $\ve_k$ determines a preimage, which determines a ball (a subset of the preimage), which determines a $\delta_k$.
Define a set of points $x_k$ by being in that preimage and different from $x_0$, i.e. take $x_k \in B^*(x_0;\delta_k)$.
Obviously, as $b_k \to \beta$, then $\delta_k \to 0$. Moreover, $x_k \to x_0$, and $f(x_k) \to \beta$.
For each $k \in \N$, we have that
$$\sup_{x \in B^*(x_0;\delta_k) \cap E} f(x) = b_k$$
and, as $\delta_k \to 0$, we have $b_k \to \beta$, giving the lim sup as desired.
(I'm sure there are details to iron out in this, but I feel there's a grain of truth at the solution in there somewhere...)
 A: We assume that in (i) $\{x_n\}$ is a sequence of points of $E\setminus\{x_0\}$, converging to $x_0$.
The statement is technical. To avoid to be lost in details, we shall follow a path, which is a suitable chosen auxiliary sequence. Namely, for each natural $n$ put $\beta_n=\sup_{x\in B^*(x_0,1/n)\cap E} f(x)\in \Bbb R\cup\{\pm\infty\}$. Clearly, a sequence $\{\beta_n\}$ is non-increasing, so there exists $\beta^*=\lim_{n\to\infty}\beta_n\in \Bbb R\cup\{\pm\infty\}$. Since the sequence $\{\beta_n\}$ is non-increasing and for each $\delta>0$ there exists natural $n$ such that $B^*(x_0,1/n)\subset B^*(x_0,\delta)$, $\beta^*=\lim\sup_{x \to x_0} f(x)$.
$(\Rightarrow)$ (i) I assume that $x_0$ is an accumulation point of the set $E$, that is $x_0\in\overline{E\setminus \{x_0\}}$, otherwise the conclusion can fail. The following case are possible:
1)) $\beta^*=-\infty$. Then for each natural $n$ we can pick $x_n\in B^*(x_0,1/n)\cap E$ such that $f(x_n)<-n$.
2)) $\beta^*=+\infty$. Then for each natural $n$ we can pick $x_n\in B^*(x_0,1/n)\cap E$ such that $f(x_n)>n$.
3)) $\beta^*\in\Bbb R$. Then for each natural $n$ we can pick $x_n\in B^*(x_0,1/n)\cap E$ such that $|f(x_n)-\beta^*|<1/n$.
(ii) Since $b>\beta^*$ and the sequence $\{\beta_n\}$ tends to $\beta^*$, there exists $n$ such that $\beta_n<b$. We can pick $\delta=1/n$.
$(\Leftarrow)$ Let $\beta<b\in \Bbb R\cup\{+\infty\}$. By (ii), $\beta_n\le b$ for all sufficiently big $n$, so $\beta^*\le b$, thus $\beta^*\le\beta$. Let $m$ be any number. Since $\{x_n\}$ is a sequence of points of $E\setminus\{x_0\}$, converging to $x_0$,
there exist a natural number $N$ such that $x_n\in B^*(x_0,1/m)$ for each $n\ge N$, so $f(x_n)\le\beta_m$. Therefore $\beta\le\beta_m$. Thus $\beta\le\beta^*$ and so $\beta=\beta^*$.
