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In the nontrivial sense, does there exist a connected subspace of $\mathbb{R}^2$ which is a union of a non-empty countable collection of closed and pairwise disjoint line segments each of unit length, i.e. length $1$? What are some good examples, if any?

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I was thinking of perhaps the set of fundamental parallelograms that tesselate the Cartesian plane, but that might not work unless we added the restriction that the sides of every parallelogram be of unit length. Hmmm… Actually it won't even work because the Cartesian plane itself is uncountable! – Libertron Nov 11 '12 at 23:49
This might work: the union of $[0,1] \times 0$ with $a/b \times [a/b,a/b+1]$ for each $a/b \in \mathbb{Q} \cap [0,1]$, where $a/b$ is written in reduced form. – Carl Nov 12 '12 at 2:28
@Carl, that's a nice idea, but I don't think it will work: for some irrational $x$, the segments $x \times [x, x + 1]$ and $[x, x + 1] \times x$ will give us a disconnection. Also, I think you meant $\mathbb{Q} \cap (0, 1]$, because $0 \times [0, 1]$ meets $[0, 1] \times 0$ at the origin. – Hew Wolff Nov 12 '12 at 5:14
@HewWolff You're right. I meant for the line segments to be $a/b \times [1/b,1/b+1]$. I'm pretty sure it doesn't work as I wrote it there. – Carl Nov 12 '12 at 20:28

I think the following works.

Let $Q_+=\mathbb Q\cap(0,1)$ and $Q_-=\mathbb Q\cap(-1,0)$. Now, define $$X = (Q_+\times[0,1])\cup([-1,0]\times Q_+)\cup(Q_-\times[-1,0])\cup([0,1]\times Q_-).$$

This is a countable disjoint union of closed unit line segments by construction. It is also connected: let $f:X\to\{0,1\}$ be a continuous map. Then $f$ is constant on each of the line segments. Let $\{q_1\}\times[0,1]$ and $\{q_2\}\times[0,1]$ be disjoint line segments. Without loss of generality, $f(q_1,0)=1$. Because $f$ is continuous, there is an open neighborhood $U$ of $(q_1,0)$ such that $f(z)=1$ for $z\in U$. Because of the connectedness of line segments, this means that there is a rational $q<0$ such that for all rationals $0>p>q$ we have $f(z)=1$ for $z\in[0,1]\times\{p\}$. Now, again because of connectedness, there is a neighborhood $V$ of $(q_2,0)$, such that $f$ is constant on $V$. But $V$ intersects at least one interval $[0,1]\times\{p\}$, where $0>p>q$. Therefore $f(q_2,0)=1$. Since $q_1$ and $q_2$ were arbitrary, this means that $f$ must be constant on $Q_+\times[0,1]$. By similar reasoning, it must be constant of the other three members of the union that defines $X$. But because these four sets are not separated, this means that $f$ must be constant on $X$ and since this means every continuous function $f:X\to\{0,1\}$ is constant, $X$ is indeed connected.

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I believe this exact construction or something very similar is in munkres as an exercise (whose point is the op's question ). – user641 Nov 12 '12 at 5:31

Consider the union of $[0,1] \times 0$ with $a/b \times [1/b,1/b+1]$ for each $a/b \in \mathbb{Q} \cap (0,1)$, where $a/b$ is written in reduced form. Let $W$ be this space. It is obvious that this set satisfies all the properties besides connectedness.

Now, assume $A,B$ are two open sets in $\mathbb{R}^2$ which induce a separation of $W$ in the subspace topology. That is, $A\cap W$ and $B \cap W$ are disjoint, and $(A \cap W) \cup (B \cap W) = W$. Clearly the component $[0,1] \times 0$ must lie entirely in either $A$ or $B$, so assume it lies in $A$. Since $A$ is open and $[0,1] \times 0$ is compact, $A$ contains a "tube" around $[0,1] \times 0$. That means that there exists some $\epsilon > 0$ such that for all $a/b$ where $0 < 1/b < \epsilon$, we have $(a/b,1/b) \in A$. But then $A$ must contain the whole connected component $a/b \times [1/b,1/b+1]$.

Now, assume $B$ is nonempty. Then it contains some component $c/d \times [1/d,1/d+1]$. In particular, it contains the point $(c/d,1)$. But then it also contains a point $(a/b,1)$ where $a/b < \epsilon$, because the rationals with denominator greater than $1/\epsilon$ are dense in $[0,1]$. But then we have $a/b \times [1/b,1/b+1] \subset B$, which is a contradiction because we already observed that this line segment must be in $A$.

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This should work, and is funny. Use polar coordinates, and identify angles with numbers in $[0,2\pi)$. For every rational $\frac ab\neq0$ in $[0,2\pi)$ with $(a,b)=1$, draw the segment $s(\frac ab)$ pointing at the origin, with angle $\frac ab$ and norm going from $\frac1b$ to $1+\frac1b$. Draw also $s(0)$ starting from the origin and going out with angle 0. Call this set $X$.

Take a clopen set $A\subseteq X$ containing $s(0)$: looking at the origin, we can say that $A$ contains every $s(\frac ab)$ for sufficiently large $b$ (because $A$ is open). But this clearly shows that $A$ is dense, hence, being also closed, $A=X$.

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I see now that my solution is not very different from Carl's one. Sorry. – Giulio Bresciani Dec 28 '13 at 21:15

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