Consider $f_\lambda (x) = \lambda x(1 - x)$ for $x, \lambda \in \mathbb{R}.$

1) Show that $K_\lambda : = \{ x \in \mathbb{R}:$ the sequence $x, f(x), f(f(x)), \ldots$ is bounded $\}$ is always compact.

2) For which values $\lambda > 0$ is $K_\lambda$ connected?

2. consider the intersection of the graphs of y=f(x) and y=x, which is a parabola and a line.

3. I managed to show 1) myself, but what do I get from this intersection?

Originally Posted by EinStone
Consider $f_\lambda (x) = \lambda x(1 - x)$ for $x, \lambda \in \mathbb{R}.$

1) Show that $K_\lambda : = \{ x \in \mathbb{R}:$ the sequence $x, f(x), f(f(x)), \ldots$ is bounded $\}$ is always compact.

2) For which values $\lambda > 0$ is $K_\lambda$ connected?
One thing that you get from the intersection of $y=x$ and $y=f_\lambda (x)$ is that for that value of x, call it $x_0$, $f_\lambda (x_0)=x_0$, in fact $f_\lambda (f_\lambda (\dots f_\lambda (x_0)\dots ))=x_0$

The vertical line, $x={1\over2}$ is the symmetry axis of the parabola $y=f_\lambda (x)$, so $f_\lambda(1-x_0)=f_\lambda(x_0)=x_0$.

Another thing to consider is that the vertex of this parabola is at $(x,\, y)=\left({1\over2},\, {\lambda\over4}\right)$.

What happens at $x={1\over2}$ if $\lambda>4$.