I have such a problem:
Prove that if $T : [0,1]\longrightarrow [0,1]$ ($T$ is continuous) is transitive (there exists $x\in X$ which has a dense orbit), then $T$ is onto, and the image of every non-degenerate subinterval of $[0,1]$ is a non-degenrate subinterval of $[0,1]$.
My problem, except the solution, is that I can't imagine this case (esspecially "onto"assumption about one dense orbit seems too weakly for me).
Solution to the second question:
We assume, that for some interval $[\alpha,\beta]\subset [0,1]$ with $\alpha<\beta$, $T([\alpha,\beta])$ is a point, say $T([\alpha,\beta])=A$ for $ A\in[0,1]$. Then there exist $x_{1},x_{2}$ such that $\alpha<x_{1}<x_{2}<\beta$ and $x_{1},x_{2}\in T^{-1}(A)$. $T$ is transitive, so there exists $x_{0}$ such that for some $m_{1},m_{2}\in\mathbb{N} \ \ \ T^{m_{1}}(x_{0})$ is arbitrarly close to $x_{1}$ and $T^{m_{2}}(x_{0})$ is arbitrarly close to $x_{2}$. Hence $T(T^{m_{1}}(x_{0})=A=T(T^{m_{2}}(x_{0})) \iff T^{m_{1}+1}(x_{0})=T^{m_{2}+1}(x_{0})$. WLOG assume $m_{1}>m_{2}$ and then $T^{m_{1}-m_{2}}(x_{0})=x_{0}$, so $x_0$ is a fixed point and we have a contradiction.