# Critical points for a smooth function defined on a manifold - Milnor, Morse theory

I need some help with a statement from Milnor's Morse theory book!

While studying the proof of a theorem I got stuck.

That's what we know: Let $f$ be a differentiable function on a manifold $M$ with no degenerate critical points, and so that for all $a \in \mathbb{R}$, $M^{a}=\{p\in M: f(p)\leq a\}$ is compact.

The author says: " Let $c_{1}<c_{2}<c_{3}<...$ be the critical values of $f:M \rightarrow \mathbb{R}$. The sequence $\{c_{i}\}$ has no cluster point since each $M^{a}$ is compact. The set $M^{a}$ is vacuous for $a<c_{1}...$"

I proved that the set of critical values of $f$ is countable and that exists a minimum. Then I showed that the mentioned sequence has non cluster point but I did not manage in showing that he set "$M^{a}$ is vacuous for $a<c_{1}...$". What I used to show the first two statement are essentially: the fact that non degenerate critical points for a smooth real valued function on a manifold are isolated(consequence of Morse's lemma), the regularity of the function $f$ and the compactness of $M$.

I thought of showing the last observation by a "proof by contradiction" supposing it exists an $a\in \mathbb{R}$ for which $M^{a}$ is not vacuous and $a<c_{1}$ and then finding a critical point for $f$: $p\in M^{a}$, leading to a contradiction. I did not really get so far in that...Do you have any hints? Suggestions?

Thanks

• If I understand correctly you have already shown that the function assumes a minimum (which is a critical point). But then there are no points with value less than the minimum, i.e. $M^a$ is vacuous for all values of $a$ below the value of the minimum. Mar 7, 2017 at 21:56
• I hadn't: I thought to show that $M^{a}$ contained a minimum but I didn't know how tho do it. Accidentally, from your previous comment, I discovered that exists a generalization of of extreme value-theorem for non empty compact topological space and so all makes sense. Thanks for the reply Mar 7, 2017 at 22:40

Let $c$ be the minimum value of $f$. Then if $f(p)=c$, all partial derivatives of $f$ must vanish at $p$ (otherwise there would be some direction you could move from $p$ to decrease the value of $p$). That is, $p$ is a critical point, so $c$ is a critical value.
So $c$ is one of the $c_i$, and since it is the minimum value of $f$, it must be $c_1$. Thus $c_1$ is the minimum value of $f$, which by definition means that $f(p)\geq c_1$ for all $p$. So if $a<c_1$, $\{p:f(p)\leq a\}$ is empty.