Placing Ts on the $x$-axis 
A "T" consists of two perpendicular intervals $\{c\}\times[0,a]$ and $[b,d]\times \{a\}$ (with $b<c<d$) on the plane. We say that the T is placed on point $c$. Is it possible to place non-intersecting T's on all real numbers on the $[0,1]$ interval of the $x$-axis?

I believe the answer should be "no". Suppose it were possible. We can assume that each T has equal left-halfwidth and right-halfwidth (i.e. half of the horizontal line of the T.) For each real number $x\in[0,1]$, let $h_x$ denote the height of its T and $w_x$ denote its halfwidth. Then two real numbers $x,y\in[0,1]$ have intersecting T's if $h_y< h_x$ and $|x-y|\le w_y$, and vice versa. So every time we have $h_x>h_y$, we must have $|x-y|>w_y$. How can we get a contradiction?
 A: Here is an analytic approach to the problem. See my other answer for a different approach.
Define a few functions: $r:[0,1]\to\mathbb R$ assigning $z$ the value $d$ of the $T$ on $z$; similarly, $l:[0,1]\to\mathbb R:z\mapsto b$.
Also, define $h:[0,1]\to\mathbb R$ by assigning $x$ the value $a$ of the $T$ at $x$. 
Note that no $x$ satisfies $h(x)=0$. Define $U=\{x\in[0,1]:h(x)<0\}$.
Theorem. $U$ is dense in $[0,1]$.
Proof. Let $(x,y)$ be an open interval in which all $T$'s are upright--$h(z)>0$ for all $z\in(x,y)$. Set $\delta=\frac{y-x}3$. Pick any point $z_0\in(x,x+\delta)$. Define inductively $z_{n+1}$ by allowing it to be any point satisfying $z_n<z_{n+1}<\min\{c_n,z_n+2^{-(n+1)}\delta\}$, where $c_n$ is the minimum of the various $r_m,m\le n$. Then $\langle z_n\rangle$ is an increasing sequence bounded above by $x+2\delta$. It converges to $z\in(x,y)$. 
Observe, $z\le r(z_n)$ for any $n$, so in order to avoid intersections $h(z)<h(z_n)$. Moreover, $z>l(z)>z_n$ for all $n$. Taking limits, $l(z)=z$, a contradiction. $\square$
Similarly, the set $V=\{x\in[0,1]:h(x)>0\}$ is also dense in $[0,1]$. We will now construct two sequences $\langle x_n\in U\rangle$ and $\langle y_n\in V\rangle$ in a fashion similar to the construction in the proof above. Pick $x_0<2^{-2}$ in $U$. Define $a_0=r(x_0)$, and choose $y_0\in V$ for which $x_0<y_0<\min\{a_0,x_0+2^{-3}\}$. Define $a_1=\min\{a_0,r(y_0)\}$. Given $x_n,y_n,a_{2n+1}$. Pick $x_{n+1}\in U$ such that $$y_n<x_{n+1}<\min\{a_{2n+1},y_n+2^{-(2n+3)}\}.$$ Let $a_{2n+2}=\min\{a_{2n+1},r(x_{n+1})\}$. Choose $y_{n+1}\in V$ such that $$x_{n+1}<y_{n+1}<\min\{a_{2n+2},x_n+2^{-(2n+4)}\}.$$ Let $a_{2n+3}=\min\{a_{2n+2},r(y_{n+1})\}$. Then, both sequences are bounded and increasing, and they have a shared limit $z$, which lies in $[0,1]$.
Since $z\le r(x_n)$ for every $n$, we can argue as in the proof of the theorem to get either $z\in V$ or $l(z)=z$. In the same way, either $z\in U$ or $l(z)=z$. Since $U\cap V=\emptyset$, it must be true that $l(z)=z$. This is a contradiction.
A: Here is a much simpler proof that there is no arrangement of non-intersecting 'T's upon $[0,1]$. This proof makes use of the fact that the cardinality of $[0,1]$ is uncountable.
For $x\in[0,1]$ let $w(x)$ be the half-width $\frac{d-b}2$ of the 'T' that contains $x$, as referred to in the question.
Given a positive integer $n$, let 
$$S_n=\left\{x\in [0,1]: w(x) \ge \frac 1n\right\}.$$
We shall make two conflicting observations. Firstly, we shall show that $|S_n|<n$ for $n>0$. Secondly, we will show that $\bigcup_{n>0}S_n = [0,1]$.
Let $n>0$, and let $x$ and $y$ be distinct elements of $S_n$. We will show $|x-y|>\frac 1n$ using a proof by contradiction. Suppose $|x-y|\le \frac 1n$. Let $h_x$ (resp. $h_y$) be the height of the 'T' that contains $x$ (resp. $y$), which is the value of $a$ in the definition provided in the question. Suppose, without loss of generality, that $h_x\le h_y$. Noting that $w(x)\ge \frac 1n$ because $x\in S_n$, it follows that $|y-x|\le w(x)$. Thus, $(y,h_x)$ belongs to the 'T' that contains $x$. Furthermore, $h_x\le h_y$, so $(y,h_x)$ belongs to the 'T' that contains $y$. Thus, the 'T' that contains $x$ intersects the 'T' that contains $y$, which is a contradiction. Therefore, $|x-y|>\frac 1n$ for all $x,y\in S_n$. Since $S_n\subset[0,1]$, we find
$$|S_n|< n.$$
Now, let $x\in[0,1]$. Then, $w(x)$ is some positive real number, by hypothesis. So, there exists some positive integer $n_0$ such that $\frac 1{n_0}\le w(x)$. So, $x\in S_{n_0}$. Since every element of $[0,1]$ belongs to some $S_n$, it follows that
$$\bigcup_{n>0}S_n = [0,1].\tag{1}$$
Now, we shall see that a contradiction arises. Indeed, the fact that $|S_n|<n$ for all $n$ implies $\bigcup_{n>0}S_n$ is countable. However, (1) implies the cardinaltiy of $\bigcup_{n>0}S_n$ is the cardinality of $[0,1]$, which is uncountable. This is a contradiction, so there is no possible arrangement of non-intersecting 'T's upon $[0,1]$.
A: This may not actually address your question, but it may help you think about it in a different way:
For any set of finitely many $x_i \in [0,1]$, we know you can construct a set of open intervals $I_i$ such that $x_i \in I_i$ and $x_i \notin I_j \;\forall j < i$. This is (should be) fairly easy to understand why; a really easy way to construct all of these intervals is if we put all our $x_i$'s in ascending order, assign the first $x$ the interval $(-\infty,\infty)$, and each subsequent $x_i$ the interval $(x_{i-1}, \infty)$. It is clear to see that we can show this for the case for any finite sequence of $x$ values.
To show how we manage this for an infinite number of values, I am going to put the sequence together like this: $\frac{1}{2}, \frac{1}{4}, \frac{3}{4}, \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}...$
We can then see that we can assign each number an interval like this: $(-\infty,\infty),(-\infty,\frac{1}{2}),(\frac{1}{2},\infty),(-\infty,\frac{1}{4}),(\frac{1}{4},\frac{3}{4}),(\frac{3}{4},\infty)...$
I'm not quite sure what this sort of sequence is called, but if we tile them this way we, in addition showing we can do this for all numbers, we get a really cool fractal structure.
This has a lout of similarity to what the problem describes, and you ay be able to do the same thing with the T's. (except for the problem that the T's have closed intervals...)
