# Sequence of critical values has no cluster point (Milnor, Morse Theory)

The following Claim is used in the proof of Theorem 3.5 in John Milnor's "Morse Theory":

Claim: Let $f: M \rightarrow \mathbb{R}$ be a differentiable function on a manifold $M$ with no degenerate critical points and such that for every $a \in \mathbb{R}$ the set $M^a=f^{-1}((-\infty, a])$ is compact. Let $c_1 < c_2 < c_3 <\ldots$ be the critical values of $f$. Then the sequence $(c_k)_{k \in \mathbb{N}}$ has no cluster point, i.e. $$\forall x \in \mathbb{R} \hspace{2mm} \exists \textit{ a neighbourhood } U \textit{ s.t. } U\cap \{c_k\}_{k \in \mathbb{N}} \textit{ is finite.}$$

The author of the book doesn't really explain why this holds, he only says that it immediately follows from the compactness of each set $M^a$. This leads me to thinking that the proof of this claim shouldn't be that difficult, however I was unable to do it.

My main approach was to prove it by contradiction. I.e. I assumed that the sequence has a cluster point $x \in \mathbb{R}$ and then for any $\epsilon > 0$ I tried to find an open cover of $M^{x+\epsilon}$ which has no finite subcover (which would contradict the compactness of $M^{x+\epsilon}$), but this didn't work.

I would greatly appreciate any help in proving the claim!

• Is Morse lemma proved already?
– user99914
Commented Apr 9, 2014 at 8:43
• Yes it is (it is Lemma 2.2 in the book). Commented Apr 9, 2014 at 8:58
• Then for each critical points $x$ there is a small neighborhood $U_x$ of $x$ such that $x$ is the only critical point in $U_x$. Then the set of all critical points are discrete. For all $c \in \mathbb R$, let $U = f^{-1}(c-1, c+1)$. This is an open set in $M$ and has compact closure. So there can only we finitely many critical points in $U$.
– user99914
Commented Apr 9, 2014 at 9:04

## 1 Answer

Corollary 2.3 claims that non-degenerate critical points are isolated, so if $$\{c_i\}$$ has a cluster point $$y,$$ we shall find critical points $$x_i\in f^{-1}(c_i)$$ such that $$\{x_i\}$$ clusters around another critical point $$x\in f^{-1}(y)$$ by compactness of $$M^{x + \epsilon}$$ and smoothness of $$f,$$ which leads to contradiction.