Let $f: {\mathbb{R}^n}$ to $\mathbb{R} $ is a continuous function. For any $x$ belonging to $\mathbb{R}^n$ define $(U(x)=\{y|f(y)>f(x)\})$ is an open set.
I tried proving that the complement of the set is closed. The property of continuity of function could be invoked. I cannot seem to properly form the proof. Please help!
On
I will prove that $U$ is sequentially open. In $\mathbb R^n$ it implies that the set is also open:
Definition (Reminder). A set $U\subseteq \mathbb R^n$ is sequentially open if, for every sequence $\{x_n\}$ converging to a point $x\in U$, there exists $n_0\in\mathbb N$ such that $x_n\in U$ whenever $n\geq n_0$.
Now, consider a point $y\in U$ and a sequence $\{y_n\}$ such that $y_n\to y$. Since $f$ is continuous, the sequence $\{f(y_n)\}$ converges to $f(y)$ on $\mathbb R$. By definition of convergence, there exists $n_0\in\mathbb N$ such that
$$f(y_n) \in (f(y)-r,f(y)+r), $$
(here $0<r<f(y)-f(x)$) whenever $n\geq n_0$ ($n_0$ may depend on $r$ but this is not important here). But this shows that $f(y_n)>f(x)$ whenever $n\geq n_0$, i.e. $y_n\in U$ whenever $n\geq n_0$, as we wanted.
$U(x)=f^{-1}((f(x),\infty)$ and $((f(x),\infty)$ is open in $\mathbb R$ so $U(x)$ is open.