Example of subsets in plane with continuous bijective mapping between them The question is from C. Pugh's Real Analysis: Construct nonhomeomorphic connected, closed subsets A, B $\subset$ $\mathbb{R}^2$ for which there exists continuous bijections $\;f: A \to B$ and $\;g: B\to A$.
I realize that $A$ and $B$ must be noncompact.
Since $A, B$ are closed $\; \Rightarrow \; A, B$ are unbounded. I am able to find $A,B\subset \mathbb R^2$ that are closed, unbounded, connected, and are not homeomorphic to each other, and then I'm stuck with the continuous bijections.
Found related examples here, but don't see how to extend them to my question.
 A: We use the usual trick. Let
$$E_0=\big(\{0\}\times\Bbb R\big)\cup\big([0,1]\times\Bbb Z\big)\cup\big((1,2)\times\{0\}\big)\,.$$
To get $E_1$ from $E_0$, replace each line segment $[0,1]\times\{n\}$ for $n\in\Bbb Z^+$ by the closed line segment from $\langle 0,n\rangle$ to $\left\langle 2-\frac1n,1\right\rangle$. To get $E_2$ from $E_1$, replace each line segment $[0,1]\times\{-n\}$ for $n\in\Bbb Z^+$ by the closed line segment from $\langle 0,-n\rangle$ to $\left\langle 1-\frac1n,-1\right\rangle$. It is easily verified that the maps
$$f_0:E_0\to E_1:\langle x,y\rangle\mapsto\begin{cases}
\langle x,y\rangle,&\text{if }y\le 0\\
\left\langle x\left(2-\frac1y\right),y-x(y-1)\right\rangle,&\text{if }y>0
\end{cases}$$
and
$$g_0:E_1\to E_2:\langle x,y\rangle\mapsto\begin{cases}
\langle x,y\rangle,&\text{if }y\ge 0\\
\left\langle x\left(2+\frac1y\right),y+x(|y|-1)\right\rangle,&\text{if }y<0
\end{cases}$$
are continuous bijections.
For $A\subseteq\Bbb R^2$ and $\langle u,v\rangle\in\Bbb R$ let $A+\langle u,v\rangle=\{\langle x+u,y+v\rangle:\langle x,y\rangle\in A\}$ and let
$$X=\bigcup_{n\in\Bbb Z^+}(E_0-\langle 2n,0\rangle)\cup E_0\cup\bigcup_{n\in\Bbb Z^+}(E_2+\langle 2n,0\rangle)$$
and
$$Y=\bigcup_{n\in\Bbb Z^+}(E_0-\langle 2n,0\rangle)\cup E_1\cup\bigcup_{n\in\Bbb Z^+}(E_2+\langle 2n,0\rangle)\,;$$
the maps
$$F:X\to Y:\langle x,y\rangle\mapsto\begin{cases}
\langle x,y\rangle,&\text{if }x<0\text{ or }x\ge 2\\
f_0(\langle x,y\rangle),&\text{if }0\le x<2
\end{cases}$$
and
$$G:Y\to X:\langle x,y\rangle\mapsto\begin{cases}
\langle x+2,y\rangle,&\text{if }x<0\text{ or }x\ge 2\\
g_0(\langle x,y\rangle)+\langle 2,0\rangle,&\text{if }0\le x<2
\end{cases}$$
are then continuous bijections.
The spaces $X$ and $Y$ are path-connected and therefore connected, and both are closed subsets of $\Bbb R^2$. They are not, however, homeomorphic.
To see this, let
$$\begin{align*}
C_X&=\{\langle 2n,1\rangle:n\in\Bbb Z^+\}\cup\{\langle 2n,-1\rangle:n\in\Bbb Z^+\}\,,\\
C_Y&=C_X\setminus\{\langle 2,-1\rangle\}\,,\text{ and}\\
F&=\{\langle 2n,0\rangle:n\in\Bbb Z\}\,;
\end{align*}$$
$C_X$ is the set of points at which $X$ is not locally connected, $C_Y$ is the set of points at which $Y$ is not locally connected, and $F$ is the set of points $p$ such that $X\setminus\{p\}$ has exactly four path components. ($F$ is also the set of points $p$ such that $Y\setminus\{p\}$ has exactly four path components.) Let $\langle x,y\rangle\in C_X$; then $\langle x,-y\rangle\in C_X$, and the obvious path in $X$ from $\langle x,y\rangle$ to $\langle x,-y\rangle$ contains exactly one point of $F$, namely, $\langle x,0\rangle$. However, $\langle 2,1\rangle\in C_Y$, and every path in $Y$ from $\langle 2,1\rangle$ to another point of $C_Y$ contains at least two points of $F$. Thus, no point of $X$ can be the image of $\langle 2,1\rangle$ under a homeomorphism, and $X$ and $Y$ are not homeomorphic.
