Let $\mathcal{I}$ be a family of closed intervals on the real line. An interval graph $G_\mathcal{I}$ is a simple graph with vertex set $\mathcal{I}$, where two vertices are adjacent if their intersection is non empty.
It is well known that when $\mathcal{I}$ (and hence $G$) is finite, a representation $\mathcal{I}'$ with $G_\mathcal{I}\approx G_{\mathcal{I}'}$ can be chosen such that for each $[a,b],[c,d]\in\mathcal{I}'$ with $[a,b]\neq[c,d]$ we have $a\neq c$ and $b\neq d$.
I want to know if this holds even if $\mathcal{I}$ is infinite?
My thoughts were that the proof for the finite case probably depends on the finiteness in the sense that whenever two endpoints coincide, we can move one interval just a tiny bit to the left or the right. It gets more complicated when more than two endpoints coincide, but this should be the basic gist.
Now if I take for example $\mathcal Q=\{[0,q]:q\in[0,1]\cap \mathbb{Q}\}$ and $\mathcal R=\{[0,r]:r\in[0,1]\}$ (or maybe also $\mathcal{Q}^\ast=\mathcal{Q}\setminus\{\{0\}\}$ and $\mathcal{R}^\ast=\mathcal{R}\setminus\{\{0\}\}$, if that changes anything), I can't see the approach to work anymore and have no idea how to proceed.
EDIT: I just noted I chose bad examples because $G_\mathcal Q\approx K_\omega$ and $G_\mathcal R\approx K_\mathfrak c$ and hence I can simply choose $\mathcal Q'=\{[-q,q]:q\in[0,1]\cap \mathbb{Q}\}$ and $\mathcal R'=\{[-r,r]:r\in[0,1]\}$. But my question remains, this may not always work. What about e.g. $\mathcal S =\{[\frac{\lfloor rn\rfloor}{n},r]:r\in[0,1]\}$ with $n$ some (big) natural number?
I think it could make a big difference whether $\mathcal{i}$ is countable or not.
If $\mathcal{I}$ is uncountable in general the statement is false. For example let $\mathcal{I}$ be the collection of all intervals on the real line. You can't rearrange them to have distinct endpoints.
If it is countable I'm not sure whether the argument for the finite case could be adapted.
Edit: I think for the countable case one can construct the graph step by step. Pick some ordering of $\mathcal{I}$. Construct the intervals by just adding them in one at a time. This is a finite problem so at each step you can find an interval with endpoints that have not been used before.