# Homeomorphism clopen sets

Does a homeomorphism between two topological spaces need to map clopen sets to clopen sets?

For example consider the Baker‘s map $$(x,y)\mapsto (2x mod 1, 1/2(y+\lfloor 2x\rfloor))$$ $$B\colon X\to X$$ on $$X=[0,1]\times [0,1]$$ with the standard topology, as well as the space shift-map $$\sigma\colon Y\to Y$$ with $$Y=\{0,1\}^{\mathbb{Z}}$$ equipped with the product topology.

Then the map $$\varphi\colon Y\to X$$ defined by

$$(\ldots,s_{-2},s_{-1},s_0,s_1,s_2,\ldots)\mapsto (x,y)=(\sum_{i=0}^\infty \frac{s_i}{2^{i+1}},\sum_{i=1}^{\infty}\frac{s{-i}}{2^i})$$

is a homeomorphism.

Now what is the image of a cylinder set, i.e. a set $$C_{a_1,…,a_k}^{n_1,…,n_k}=\{(s_i)\in Y: s_{n_j}=a_j: j=1,…,k \}$$

Such cylinder sets are clopen but its Image cannot be clopen since X does not have clopen sets in the standard topology (despite of X and the emptyset).

But since $$\varphi^{-1}$$ is also continuous, the Image of a cylinder set needs to be clopen!

• Why do you think that $\varphi$ is a homeomorphism? It cannot be exactly for the reason you pointed out. (Or simpler because $Y$ is disconnected while $X$ is connected) Mar 3, 2022 at 13:08
• But the Bakers map and the Shift are topologically conjugated, aren‘t they? Mar 3, 2022 at 13:11
• Some authors use topologically conjugate to mean that there exist a continuous injection $\varphi\colon Y\to X$ with $\varphi\circ\sigma=B\circ\varphi$, without requiring $\varphi$ to be a homeomorphism, maybe this is where the confusion is coming from Mar 3, 2022 at 13:20
• But the map $\varphi$ is bijective... where does it fail to be a homeomorphism? Mar 3, 2022 at 15:09
• Continuous bijections are not necessarily homeomorphisms. Indeed, the inverse may fail to be continuous, which is precisely what is needed to preserve clopen sets. Mar 3, 2022 at 15:23

So if your $$\phi$$ does not, it cannot be a homeomorphism.
$$Y$$ is (totally) disconnected and $$X$$ is connected so there cannot be a homeomorphism between them.
• I see. I would like to understand which properties of a homeomorphism the function $\varphi$ has (and which not). As far as I see, $\varphi$ is continuous and bijective. Then, its inverse $\varphi^{-1}$ cannot be continuous. Is there a way to see this without using that, otherwise, one would have a homeomorphism which is impossible due to the conectedness/ discommectedness? Mar 3, 2022 at 20:08