The answerThis is not an answer but an extended comment to indicate that the answer would be no without the simplicity or convexity assumptions. First observe that there are curves in $\textbf{R}^2$ with exactly $4$ critical points of curvature which self-intersect an unlimited number of times, as shown for instance in the picture below.
Pull this curve to $\mathbf{S^2}$ via stereographic projection, which will preserve critical points of curvature. So we obtain a spherical curve whose torsion changes sign exactly $4$ times and intersects some plane more than $4$ times. AThis example is convex but not simple. A perturbation of it yields a simple closed curve in $\mathbf{R}^3$ which still has exactly $4$ vertices butand intersects some plane more than $4$ times, but this perturbation will not preserve convexity.
The answer is no. First observe that there are curves in $\textbf{R}^2$ with exactly $4$ critical points of curvature which self-intersect an unlimited number of times, as shown for instance in the picture below.
Pull this curve to $\mathbf{S^2}$ via stereographic projection, which will preserve critical points of curvature. So we obtain a spherical curve whose torsion changes sign exactly $4$ times. A perturbation yields a simple closed curve in $\mathbf{R}^3$ which has exactly $4$ vertices but intersects some plane more than $4$ times.
This is not an answer but an extended comment to indicate that the answer would be no without the simplicity or convexity assumptions. First observe that there are curves in $\textbf{R}^2$ with exactly $4$ critical points of curvature which self-intersect an unlimited number of times, as shown for instance in the picture below.
Pull this curve to $\mathbf{S^2}$ via stereographic projection, which will preserve critical points of curvature. So we obtain a spherical curve whose torsion changes sign exactly $4$ times and intersects some plane more than $4$ times. This example is convex but not simple. A perturbation of it yields a simple closed curve in $\mathbf{R}^3$ which still has exactly $4$ vertices and intersects some plane more than $4$ times, but this perturbation will not preserve convexity.
The answer is no. First observe that there are curves in $\textbf{R}^2$ with exactly $4$ critical points of curvature which self-intersect an unlimited number of times, as shown for instance in the picture below.
Pull this curve to $\mathbf{S^2}$ via stereographic projection, which will preserve critical points of curvature. So we obtain a spherical curve whose torsion changes sign exactly $4$ times. Perturbing this curve we obtainA perturbation yields a simple closed curve in $\mathbf{R}^3$ which has exactly $4$ points of vanishing torsionvertices but intersects some plane more than $4$ times.
The answer is no. First observe that there are curves in $\textbf{R}^2$ with exactly $4$ critical points of curvature which self-intersect an unlimited number of times, as shown for instance in the picture below.
Pull this curve to $\mathbf{S^2}$ via stereographic projection, which will preserve critical points of curvature. So we obtain a spherical curve whose torsion changes sign exactly $4$ times. Perturbing this curve we obtain a simple closed curve in $\mathbf{R}^3$ which has exactly $4$ points of vanishing torsion but intersects some plane more than $4$ times.
The answer is no. First observe that there are curves in $\textbf{R}^2$ with exactly $4$ critical points of curvature which self-intersect an unlimited number of times, as shown for instance in the picture below.
Pull this curve to $\mathbf{S^2}$ via stereographic projection, which will preserve critical points of curvature. So we obtain a spherical curve whose torsion changes sign exactly $4$ times. A perturbation yields a simple closed curve in $\mathbf{R}^3$ which has exactly $4$ vertices but intersects some plane more than $4$ times.
The answer is no. First noteobserve that we can construct a curvethere are curves in $\textbf{R}^2$ with exactly $4$ critical points of curvature which intersects itselfself-intersect an unlimited number of times, as shown for instance in the picture below.
Pull this curve to $\mathbf{S^2}$ via stereographic projection, which will preserve critical points of curvature. So we obtain a spherical curve whose torsion changes sign exactly $4$ times. Perturbing this curve we obtain a simple closed curve in $\mathbf{R}^3$ which has exactly $4$ points of vanishing torsion but intersects some plane more than $4$ times.
The answer is no. First note that we can construct a curve in $\textbf{R}^2$ with exactly $4$ critical points of curvature which intersects itself an unlimited number of times, as shown for instance in the picture below.
Pull this curve to $\mathbf{S^2}$ via stereographic projection, which will preserve critical points of curvature. So we obtain a spherical curve whose torsion changes sign exactly $4$ times. Perturbing this curve we obtain a simple closed curve in $\mathbf{R}^3$ which has exactly $4$ points of vanishing torsion but intersects some plane more than $4$ times.
The answer is no. First observe that there are curves in $\textbf{R}^2$ with exactly $4$ critical points of curvature which self-intersect an unlimited number of times, as shown for instance in the picture below.
Pull this curve to $\mathbf{S^2}$ via stereographic projection, which will preserve critical points of curvature. So we obtain a spherical curve whose torsion changes sign exactly $4$ times. Perturbing this curve we obtain a simple closed curve in $\mathbf{R}^3$ which has exactly $4$ points of vanishing torsion but intersects some plane more than $4$ times.