Tent map is topologically transitive Let $T:[0,1] \to [0,1]$ be the function $Tx= 2x$, if $ x \in [0, \frac{1}{2}]$ and $Tx = 2-2x$, if $ x \in ( \frac{1}{2} , 1] $. We say that a map is topologically transitive if, for any pair $U, V$ of nonempty open subsets of $X$, there exists some $n\ge0$ such that $T^{n}(U) \cap V \neq \varnothing$. I need to show that $T$ is topologically transitive.
My attempt was to show that for any open and nonempty subset of $X$, say $U$, there exists $n\ge0$ such that $T^n(U)=[0,1]$ and then for any open and nonempty subset of $X$, say $V$, $T^n(U)\cap V=[0,1]\cap V=V\neq \varnothing$. What I have tried so far is to do this by cases:
First I suppose $U$ is an open interval containing $\frac{1}{2}$. Observe that $T(U)$ contains $T(\frac{1}{2}) = 1$ and $T^{2}(U)$ contains $T(T( \frac{1}{2}))=T(1)=0$, so $T^{2} (U)$ is an interval which contains the number $0$, so $[0,a]\subseteq T^{2}(U)$ for some $a\in [0,\frac{1}{2}]$. We also observe that $T$ is doubling the length of the interval $[0, \frac{1}{2}]$ as we iterate it. So, there exists $n\ge0$ such that $T^{n+2}(U)$ is an interval which contains $0$ and $\frac{1}{2}$ and then $T^{n+3}(U)$ contains $0,1$ which show that $T^{n+3}(U)=[0,1]$.
The second case is if $U$ is an open subinterval of $[0,\frac{1}{2}]$. In this case we observe that $T$ is doubling the length of the interval, as we iterate it and it gets to a point that there exists some $n\ge0$ such that $\frac{1}{2}\in T^n(U)$, so from the first case there exists some $m\ge0$ such that $T^m(T^n(U))=[0,1]\implies T^{m+n}(U)=[0,1]$.
The third case is that $U=(a,b)\subseteq(\frac{1}{2},1)$, then $T(U)=(2-2b, 2-2a)$. We observe that $T(U)$ is double the length of the interval $(a,b)$. I want to show that there exists a $n\ge0$ such that $\frac{3}{4} \in T^n(U)$ and then $T(\frac{3}{4})=\frac{1}{2}\in T^{n+1}(U)$, so I apply the first case again. This is where I got stuck and I don't know how to continue. I would appreciate any help with this, or any other kind of solution.
 A: First we need, as you suspected and attempted, the lemma:

$I\subseteq[0,1]$ is a nontrivial closed interval, $T^n(I)=[0,1]$ eventually.

Core to the proof is that the continuous image of an interval is an interval.
Proof:

Suppose $0\in I$, so $I=[0,a]$ for some $a$. For some $n$ we will then have, for some $x$, $T^n(I)=[0,x]\supseteq[0,1/2]$ and then $T^{n+1}(I)=[0,1]$. Now suppose $I\subseteq(0,1/2]$. After some number of iterations $n$ clearly we will have either $T^n(I)$ straddling $1/2$ or $T^n(I)\subseteq[1/2,1]$. We treat the latter case last.
For the former, let $T^n(I)=[a,b]$, and then the hypothesis is that $a\lt\frac{1}{2}\lt b$. $T^{n+1}(I)$ will then equal $[x,1]$ for some $x$; $T^{n+2}(I)$ will equal $[0,y]$ for some $y$ and we have already covered this case.
Finally suppose $I=[a,b]$,$1/2\lt a\lt b\lt1$ (the case when the interval contains $1$ has already been covered, and if the interval contains $1/2$ its next time iterate contains $1$). Our dream is to get this interval into one of the already seen forms - this dream fails if and only if $T^n(I)$ remains in the above form for all $n$. Let’s analyse this failure: consider that $1/2\lt T(x)\lt 1$ and $1/2\lt x\lt1$ implies further that $1/2\lt x\lt3/4$. What if $1/2\lt T^2(x)\lt1$ as well? Then $1/2\lt T(x)\lt3/4$ and $5/8\lt x\lt3/4$ - the possibilities for $x$ are shrinking.
These numbers come from solving $2-2x=y$, which is testing the assumption that for all $x\in I$, the orbit of $x$ is strictly between $1/2$ and $1$ so that the action of $T$ is always $x\mapsto2-2x$ on this orbit. $2-2x=y\implies x=1-y/2$, and thus the preimage of any interval $(c,d)$ by this map is $(1-d/2,1-c/2)$ which has length $(d-c)/2$. Thus the infinite intersection of preimages can have at most one point ($2/3$, in fact), since the diameter vanishes ($1/2,1/4,1/8,\cdots\to0$). However, as a nontrivial interval, $I$ cannot have only one point, so it is impossible that every element of $I$ should have the entirety of its orbit contained in $(1/2,1)$.
Accordingly, our dream never fails, and $I$ evolves over time into one of the already-covered cases. The event that $I$ gets caught in a loop, i.e. always going from the case $I\subset(1/2,1)$ to the case $I\subset(0,1/2)$ and back, is impossible by similar argument. $\blacksquare$

With this lemma in hand, note that if $U,V$ are any two nonempty open subsets of $[0,1]$ then $U$ must contain a point $x_0\in[0,1]$ and an open interval $(x_0-r,x_0+r)$ for some $r\gt0$, and it must contain the nontrivial closed interval $I:=[x_0-r/2,x_0+r/2]$. Using the lemma, as: $$\emptyset\neq V\cap[0,1]=V\cap T^n(I)\subseteq V\cap T^n(U)$$For some $n\in\Bbb N$, the claim holds and we are done by the arbitrariness of $U,V$.
