Isolatedness of critical points of Morse functions

Question

Suppose $f:\mathbb{R}^n\to\mathbb R$ is a Morse function , i.e. is smooth and all its critical points are non-degenerate . Intuitively, why all the critical points of $f$ happen to be isolated? How would you prove this without mentioning manifolds ?(Morse theory has applications in classifying manifolds)

Answer 1

By Morse lemma, any non-degenerate critical point is locally on the form $$ (x_1, \dots, x_n) \mapsto -x_1^2 + \dots - x_r^2 + x_{r+1}^2 + \dots + x_{n}^2$$ where $r$ is the number of negative eigenvalues of the Hessian matrix. It's clear that $0$ is the only critical point in a small neighbourhood.

You don't need manifold, just implicit function theorem and Hadamard's lemma are enough, for example looks here (don't be misguided by the title, the proof actually does not use any manifold !)