This is follow up question of the following two questions I asked
I would like to show that there is $\limsup_{x\to p}f(x)$ is the unique real number, which satifies:
Proof:
Suppose there is $c\in\mathbb{R}$ s.t. conditions $a)$ and $b)$ are satisfied and $c\neq\limsup_{x\to p}f(x)$. W.l.o.g. suppose $\limsup_{x\to p}f(x)>c$
By condition $a)$ we have that there is $\tilde{\delta}$ such that $\forall x\in B_{\tilde{\delta}}(p): f(x)<c$.
But since $c$ satisfies both $a)$ and $b)$ too, it follows by condition $b)$ that
$$\begin{align}\forall\delta>0\exists x_{\delta}\in B_{\delta}(p): f(x_{\delta})>c \end{align}$$
Thus, for $\tilde{\delta}$ there is $x_{\tilde{\delta}}$ s.t. $f(x_{\tilde{\delta}})>c$.
$x_{\tilde{\delta}}\in B_{\tilde{\delta}}(p)$. Therefore $f(x_{\tilde{\delta}})<c$.Therefore
$$\begin{align}c<f(x_{\tilde{\delta}})<c\end{align}$$, which is a contradiction. Hence, $\limsup_{x\to p}f(x)$ is the unique element satisfying $a)$ and $b)$.
