On the existence of a nontrivial connected subspace of $\mathbb{R}^2$

Solution 1:

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.

Solution 2:

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 $A$.