Let f be a smooth function with $f: Ω⊂R^m→R$ Is the set $\{(x,y)∈Ω×R|y=f(x)\}$ open?
A set is considered open if, for every point within the set, you can find a small enough region around that point that is entirely contained within the set.
I am thinking that the set is open since in our case, for any point $(x, f(x))$ on the graph, you can find a small enough neighborhood around it (in the domain $Ω$) such that all the points in that neighborhood, when combined with the corresponding y values (i.e., $f(x)$), lie within the graph.
Am I right that the set is open? Its not really mathematically shown since I argued in a more informal way.
You can solve this problem using the significantly weaker assumption that $f$ is continuous. As $\mathbb{R}$ is Hausdorff, it follows that the graph $G_f=\{(x, y) \in \Omega \times R : f(x) = y\}$) is closed in $\Omega \times R$ (see this article on Wikipedia and the reference on the sufficient conditions section for more details).
Showing that $G_f$ is closed is not sufficient in showing that it is not open, so we'll now prove that indeed it is indeed not open: we'll argue by contradiction, so suppose that $G_f$ is clopen. Let $x \in \Omega$ be an arbitrary point and consider its connected component $X \subseteq \Omega$. As $X$ and $R$ are connected, so is their product $X \times R$. Restricting $G_f$ to the subspace $X \times R$, we obtain the clopen set $G_f \cap (X \times R)$, which together with the fact that $X \times R$ need be connected, implies that either $G_f \cap (X \times R) = \emptyset$ or that $G_f \cap (X \times R) = (X \times R)$, both cases being visibly nonsense.
I don't think that some sort of informal reasoning is feasible in solving a technical problem as such, if you want to read more on topology, I wholeheartedly recommend reading Mendelson's introduction to topology as a first book on the topic - I personally found it very "user-friendly" when I was younger.