I don't know if it matters, but let us assume $M$ is a smooth manifold without (manifold) boundary, which is symbolized by $\partial M=\emptyset$. Now suppose $f:M\to\Bbb R$ is a smooth function on $M$. Given $a\in\Bbb R$, I was wondering if it is possible to prove or disprove that the (topological) boundary $bdy(\{x:f(x)<a\})$ of the strict sub-level set $\{x:f(x)<a\}$ is equal to $\{x:f(x)=a\}$.
Edit 1: Since $f$ is smooth, it is continuous, which shows $\{f<a\}$ is open. Then $bdy(\{f<a\})$ is equal to the difference of the closure of $\{f<a\}$ and $\{f<a\}$. In symbols, we write $$bdy(\{f<a\})=\overline{\{f<a\}}\setminus\{f<a\}.$$ Now it is tempting to show that $\overline{\{f<a\}}=\{f\leq a\}$. But when I was doing it, all that came into my mind was to express $\{f<a\}$ as
Does it help any? This is a countable union of closed sets, so there seems to be no point in saying that the union is closed. And I'm stuck here. Does anyone have an idea? Thank you.
Edit 2: Thanks to @AnCar, we have a counter-example (see the comment). Now, to make the equality come true, we might have to require that the gradient $\mathrm{grad\ } f$ of $f$ should not vanish somewhere, which presumes $M$ to be endowed with a Riemannian metric $g$. So let me update the question as follows. Given a smooth real-valued function $f$ on a Riemannian manifold $(M,g)$ without boundary, can we prove or disprove $$bdy(\{f<a\})=\{f=a\}$$ if $\mathrm{grad\ } f\neq 0$ everywhere? To prove $bdy(\{f<a\})\subseteq\{f=a\}$, we begin by randomly selecting $$x_0\in\overline{\{f<a\}}\cap\{f\geq a\}=bdy(\{f<a\}).$$ Since $f(x_0)\geq a$, we can divide the proof into two parts. The first one comes to us when we consider $f(x_0)=a$, which makes certain of the inclusion. So we proceed to consider the second part, in which $f(x_0)>a$. Now, $f$ is continuous, so we can find a neighborhood $U$ of $x_0$ so that $f>a$ on $U$. Clearly, $U$ does not intersect $\{f<a\}$, which makes $x_0$ unable to fall into the closure, a contradiction. Now we can move on to determine if the other inclusion is true: $$\{f=a\}\overset{\color{red}?}{\subseteq}bdy(\{f<a\})$$ This is where I get lost. If $(M,g)$ is simply the real axis $\Bbb R$ with the standard metric, I can obtain the desired result, though. In this case, whenever $x_0\in\{f=a\}$, every open ball centered at $x_0$ will contain points where $f<a$ and points where $f>a$ because $f'(x_0)\neq 0$. This implies $x\in bdy(\{f<a\})$. Now, how can I generalize this argument to an abstract Riemannian manifold $(M,g)$, especially when there is this bugaboo $\mathrm{grad\ } f\neq 0$? I need some help, please. By the way, does my question have to do with regular values? Thank you.
I am not a differential geometry guy, so I don't feel comfortable writing this in the language of Riemannian manifolds. I will just assume that $M$ is a smooth manifold and that $f:M\to\mathbb{R}$ is a differentiable function with non-zero gradient.
For arbitrary $a\in\mathbb{R}$, set $M_{a,-}:=\{x\in M: f(x)<a\}$, $M_{a,+}:=\{x\in M: f(x)>a\}$ and $M_a:=\{x\in M: f(x)=a\}$. The question then becomes to show that $\overline{M_{a,-}}\setminus M_{a,-}^\circ=M_a$.
Let us also observe that $M_{a,-}^\circ=M_{a,-}$ simply because $f$ is continuous and $M_{a,-}$ is the preimage of the open set $(-\infty,a)$ through the continuous map $f$.
In order to show the first inclusion `$\subseteq$', it therefore suffices to show that $\overline{M_{a,-}}\cap M_{a,+}=\varnothing$. This is straightforward, as already showed in the original post. If $x\in \overline{M_{a,-}}\cap M_{a,+}$, then for $y$ in a small neighborhood of $x$ we would have that $f(y)>a$, meaning that $x$ cannot be a limit point of $M_{a,-}$.
We are left to show the other inclusion `$\supseteq$'. For this, since clearly $M_a\cap M_{a,-}= \varnothing$, we just need to argue that $M_a\subseteq \overline{M_{a,-}}$. So let $x\in M_a$ be arbitrary, i.e., let $x\in M$ such that $f(x)=a$ be arbitrary. We need to show that for any open neighborhood $U\subset M$ containing $x$, we must have that $U\cap M_{a,-}\neq\varnothing$, i.e., that there exists some $y\in U$ such that $f(y)<a$.
So let us assume that there exists an open neighborhood $U$ of $x$ such that $f(y)\geq a$ for all $y\in U$. Note that this property will be preserved for any included neighborhood that still contains $x$. So, by possibly relabeling $U$, we can apply the definition of a smooth manifold to find a diffeomorphic chart map $\phi:B\to U$ satisfying $\phi(0)=x$, where $B$ is the unit open ball in $\mathbb{R}^d$, $d$ being the manifold dimension of $M$.
Note that $(f\circ\phi)(B)\subset[a,\infty)$ by our assumption on $U$. However, $(f\circ\phi)(0)=a$, so $0$ is a minimum of the differentiable map $f\circ\phi$. Therefore, $\nabla (f\circ\phi)(0)=0$. By the chain rule, since $\phi$ is a diffeomorphishm, this means that $\nabla f$ must have a $0$, contradiction.