I got a trouble in proving the iff statement of lim sup.
The following is what I got from my homework.
Let $X\subseteq \Bbb R^m$, $f:X\to\Bbb R$ be a function, and $y$ be a limit point of $X$.
It is said that $L$ is the limit superior of $f$ at $y$, defined by
$\overline \lim_{x\to y}f(x)=\lim_{\delta \to 0^+} (\sup \{f(x):x\in E \cap B(y,\delta)-\{y\}\})$.
The problem I need to work is to verify the following assertions:
1) For all $L \in \Bbb R, \overline \lim_{x\to y}f(x)\le L \iff \forall \varepsilon\gt0\ \exists \delta \gt0\ \forall x\in X\cap B(y,\delta)-\{y\},\ f(x)\le L+\varepsilon$.
2)For all $L \in \Bbb R, \overline \lim_{x\to y}f(x)\ge L \iff \forall \varepsilon\gt0\ \forall \delta \gt0\ \exists x\in X\cap B(y,\delta)-\{y\},\ f(x)\ge L-\varepsilon$.
The following was what I tried on the first assertion:
Suppose $\lim_{x \to y}f(x)=l\le L$. Then, by the definition of limit, I will have:
$\forall \varepsilon \gt0\ \exists \delta\gt0\ \forall x \in X\cap B(y,\delta)-\{a\}, \ |\sup f(x)-l|\lt \varepsilon$. That is, $\sup f(x) \lt l+\varepsilon$.
Since $\forall x \in X\cap B(y,\delta)-\{y\}$, I will have $f(x)\le\sup f(x)$, then $f(x)\le\sup f(x)\lt l+\varepsilon \le L+\varepsilon$.
I do not whether this attempt is on the right track since, strictly speaking, I can only get $f(x)\lt L+\varepsilon$.
And I have no idea about the second assertion.
For convenience, let's call $X_\delta = E \cap B(y,\delta)\setminus \{y\}$.
1) Suppose $\overline\lim_{x\to y}f(x) = l \leq L$, and let $\varepsilon> 0$. Then there exists $h_\varepsilon > 0$ such that if $0 < \delta < h_\varepsilon$, then $\sup\{f(x): x\in X_\delta\}-l < \varepsilon$. We don't need absolute value here: as $\delta \to 0^+$ $\sup\{f(x):x\in X_\delta\}$ is decreasing, so $l \leq \sup\{f(x):x\in X_\delta\}$ and therefore $|\sup\{f(x):x\in X_\delta\}-l| = \sup\{f(x):x\in X_\delta\}-l$. Then for all $x \in X_\delta$, we have $f(x) \leq \sup \{f(x):x\in X_\delta\} < l+\varepsilon \leq L+\varepsilon$. So our desired "$\delta$" is $h_\varepsilon$. The fact that I got $<$ instead of $\leq$ is because I insisted on $\sup\{f(x): x\in X_\delta\}-l < \varepsilon$; of course it's completely equivalent to have $\sup\{f(x): x\in X_\delta\}-l \leq \varepsilon$.
For the reverse direction, suppose the RHS holds. We want to show that $\lim_{\delta\to 0^+}\sup\{f(x): x\in X_\delta\} = l \leq L$. Suppose not. Then $l > L$. Since $\sup\{f(x): x\in X_\delta\}$ decreases as $\delta \to 0^+$, we must have $\sup\{f(x): x\in X_\delta\}\geq l > L$ for all $\delta > 0$. Take $\varepsilon = l -L$. For all $\delta > 0$, we have $\sup\{f(x): x\in X_\delta\} - L \geq l - L > \varepsilon$, i.e. $\sup\{f(x): x\in X_\delta\} > L+\varepsilon$. Then there must exist $x \in X_\delta$ such that $f(x) > L+\varepsilon$. (If the supremum is larger than something, then there must be something in the set between that thing and the supremum.) In summary: $(\exists \varepsilon > 0)(\forall \delta > 0)(\exists x \in X_\delta) f(x)>L+\varepsilon$. This contradicts the RHS, and the proof is complete.
2) Suppose $\overline\lim_{x\to y}f(x) \geq L$. Since $\sup \{f(x):x\in X_\delta\}$ is decreasing as $\delta \to 0^+$, we must have $\sup\{f(x):x \in X_\delta\} \geq L$ for all $\delta > 0$. Let $\varepsilon > 0$ and $\delta > 0$. Then $\sup\{f(x):x \in X_\delta\} \geq L > L - \varepsilon$. Then there exists $x \in X_\delta$ such that $\sup\{f(x):x \in X_\delta\} > f(x) \geq L-\varepsilon$, and we are done.
For the reverse direction, assume the RHS and suppose (aiming for contradiction) $\lim_{\delta\to 0^+}\sup\{f(x):x\in X_\delta\} < L$. Take $\varepsilon < L-\lim_{\delta\to 0^+}\sup\{f(x):x\in X_\delta\}$. Then $\lim_{\delta\to 0^+}\sup\{f(x):x\in X_\delta\} < L-\varepsilon$. Then for some $\delta > 0$, we have $\sup\{f(x):x \in X_\delta\} < L-\varepsilon$. So for all $x \in X_\delta$, $f(x) < L-\varepsilon$. Thus $(\exists \varepsilon > 0)(\exists \delta > 0)(\forall x\in X_\delta) f(x)<L-\varepsilon$, contradicting the RHS.