On the existence of a nontrivial connected subspace of $\mathbb{R}^2$ 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?
 A: 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.
A: 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 $A$.
A: 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, $A=X$.
