Let $(X,\tau)$ be a topological manifold and $f:X\to\Bbb R$ a smooth function. I need to know if the support of $f$, defined as $\text{supp}(f)=\{x\in X\mid f(x)\neq 0\}$ is open in $\tau$. I think I have a proof but I'm not sure about its veracity.
Let $x$ be a point on the boundary of $\text{supp}(f)^\complement=\{x\in X\mid f(x)=0\}$. If $f(x)=k\neq0$ we have that for all points $p\in\text{supp}(f)$ arbitrarily close to $x$, $k\leq d(f(x),f(p))$, but this contradicts that it is smooth, so $\text{supp}(f)$ must contain its boundary, meaning that $\text{supp}(f)$ is equal to its closure , which means it is closed, meaning that $\text{supp}(f)$ is open. Is this correct?
Best Answer
Maybe I'm missing something here but since $\mathbb{R}$ is a Hausdorff space it would be enough even if $f$ was just continuous. $\mathbb{R}\setminus\{0\}$ is an open set in $\mathbb{R}$ so its inverse image must be open in $X$.