Can the plane be covered by open disjoint one dimensional intervals? I remember I heard this question a while ago but have never figured it out. Let an open interval in $\mathbb{R}$ of the form $(a,b)$. Now, imaging placing it in $\mathbb{R}^2$, anywhere we want while keeping its length but perhaps rotating it if we want. Essentially this creates a line segment without its start/endpoints parametrized by
$P(1-t)+Qt$ for $0<t<1$ where $P$ and $Q$ are its start and end points and with length $||P-Q||=b-a$.
As an example, if we take the unit interval $(0,1)$, we can place it as it is in $\mathbb{R}^2$ as $(0,1)\times \{0\}$ or we can rotate it as $\{1\}\times (0,1)$ etc. I hope it is clear what I mean. My question is the following.
Can we cover the entire $\mathbb{R}^2$ with disjoint such "open intervals" or if you prefer, line segments of finite length without their start/endpoints?
I don't remember if the initial question I read was using only unit intervals, or of any length $(a,b)$ and I don't know if that plays a role in the answer but as a gut feeling, I doubt it does. My impression is that the answer is negative and we can't cover the plane in such a manner, or if it is possible it would use some weird abstract construction perhaps involving AC. Any help is appreciated!
Edit: It seems that I could have phrased my wording a bit better although that was my last paragraph in the post above that I wasn't certain if it mattered whether we need to have fixed length intervals or not. Apparently it does, it is impossible with using only fixed length intervals but it is possible if we allow intervals of different lengths. Proof of impossibility in one case given in a paper given in the answers, and a construction in the other case is given by some answers and also found in that paper itself. Thanks everyone!
 A: Yes, it can be done.  Here is a recursive construction that uses only axis-aligned segments.  Start by covering the open unit square $(0,1)\times(0,1)$ with horizontal segments.  Then, given an initial open square, make 8 more copies and arrange them around the initial square, producing a larger open square with an uncovered hash symbol or octothorpe (#).  Then cover the octothorpe with two long horizontal segments and six short vertical segments.  You’ve now tripled the size of the initial open square.  Repeating this indefinitely covers all of $\mathbb{R}^2$.
It is an interesting question whether there is a covering using segments that are all the same length.  The best I see right now is a covering using segments of length $1$ and countably many segments of length $2$.
A: In fact it is impossible with open intervals, although you can do it with closed intervals:
https://mathoverflow.net/questions/43611/decomposing-the-plane-into-intervals
A: My guess is that you can do it.
$\qquad\qquad\qquad$
Notice first that you can tile the plane with open (rectangular) triangles and open squares. In  fact, grid the plane with rotated/translated copies of the segment $(0,1)$, obtaining a regular lattice of open squares (notice you have still to cover the endpoints of the segments).
Now add diagonal segments of length $2\sqrt 2$ in order to cut one square every 2 into 4 rectangular triangles. You can do it by shifting every parallel diagonal by $\sqrt 2$.
But now you can cover any open triangle (that is, without its perimeter) and open square with segments parallel to one of the sides.
A: Assuming that the intervals can be of different lengths, I think it can be done. I sketched something on a paper and it seems to me like this is a solution that only uses 2 different lengths for the intervals, namely $1$ and $\frac 12$:
First, create a fabric-looking structure from horizontal intervals of the form $(m, m+1) \times \{k\}$ and $\left(m+\frac 12, m+\frac 32\right) \times \left\{k+\frac 12\right\}$ and vertical ones of the form $\{k\} \times \left(m+\frac 12, m+\frac 32\right)$ and $\left\{k+\frac 12\right\}\times(m,m+1)$, $\forall k,m\in\mathbb{Z}$. Then just complete it with parallel segments of length $\frac 12$, for example, with all-vertical segments: $\{k+r\}\times\left(m,m+\frac 12\right), \{k+r\}\times\left(m+\frac 12, m+1\right)$, $\forall k,m \in\mathbb{Z}, r\in(0,1)\setminus\left\{\frac 12\right\}$.
I would have added an illustration, but all I have is a pen-on-paper sketch and I'm using a phone. I can add a photo of that if requested.
