Suppose $F(x,t): X\times I \rightarrow R$ is a homotopy of Morse functions. That is, $f_t: X \rightarrow R$ is Morse for every $t$. Show that the set $C = \{(x,t)\in X\times I : d(f_t)_x = 0\}$ forms a closed, smooth submanifold of dimension one of $X\times I$. Assume the homotopy is constant near the ends of I and use an open interval.
What I have done so far is that showed it is a smooth manifold by inverse function theorem. But I don't know how to show it is closed and has dimension 1.
Thank you very much for your guidance!
Thank you for @DanielMcLaury's comment, I finally figured it out after his very wise guidance.
Submanifold: When $dF=0, d^2 F \neq 0$, we know the Jacobian of $dF$ is not singular. Hence by inverse function theorem, it is a submanifold.
Closed: The given condition {$(x,t) \in X \times I:d(f_t)_x=0$} is closed, since all the limit points satisfy the condition and therefore is included in the set.
1-dimension: By Morse function condition, because $d^2 F \neq 0$, that means given $df_x = 0$, that means $df_{x^\prime} \neq 0$ for all $x^\prime$ in any neighborhood of $x$. Therefore, the set can be only varied by $t$, which suggests it is 1-dimensional.