Non-homeomorphic connected one-dimensional Hausdorff spaces that have continuous bijections between them in both sides

Question

I need to construct an example of two non-homeomorphic connected one-dimensional Hausdorff spaces that have continuous bijections between them in both sides. Spaces should have induced ("good") topology. I know that there does not exist an appropriate example among graphs with finite number of vertices and edges. I could not find the desired spaces here: Non-homeomorphic spaces that have continuous bijections between them and anywhere else. Please, help.

Answer 1

Here's such a construction, actually producing an infinite family of such spaces, actually planar and locally compact, pairwise in continuous bijection in both directions but pairwise non-homeomorphic.

Let $w:\mathbf{Z}\to\{0,1\}$ be a sequence. Let $\mathcal{W}$ be the set of such sequences.

For $w\in\mathcal{W}$, let $X_w$ be a planar set defined as the union of the line $\mathbf{R}\times\{0\}$ and, attached at each $(n,0)$, $n\in\mathbf{Z}$:

• if $w(n)=1$: a little half-open interval, say $\{(n,t):0\le t<1\}$
• if $w(n)=0$, a little topological circle, say the circle centered at $(n,1/4)$ of radius $1/4$.

For $w,w'\in\mathcal{W}$, say that $w'$ is a monotone reindexing of $w$ if $w'(n)=w(\varepsilon n+c)$ for some $c\in\mathbf{Z},\varepsilon\in\{1,-1\}$ and all $n\in\mathbf{Z}$.

Then for $w,w'\in\mathcal{W}$,

• $X_w$ is homeomorphic to $X_{w'}$ if and only if $w'$ is a monotone reindexing of $w$;
• $X_w$ has a continuous bijection onto $X_{w'}$ iff there is a monotone reindexing $w''$ of $w'$ such that $w\ge w''$.

Let $\mathcal{W}_0$ be the (infinite countable) set of $w\in\mathcal{W}$ such that

• $w(n)=0$ for all $n<0$
• $w(0)=1$
• $w(n)=1$ for all $n\gg 1$.

Then from the above, it follows that for all $w,w'\in\mathcal{W}_0$, there is a continuous bijection $X_w\to X_{w'}$, while there they are homeomorphic if and only if $w=w'$.

(Possibly one can also find such a subset consisting of uncountably many sequences, but I haven't tried to solve this combinatorial exercise, since OP is just asking for a pair of spaces.)