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

312 Views Asked by At

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!

1

There are 1 best solutions below

0
On

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.