# Decomposing $\{0,1\}^\omega$ endowed with the Sierpinski topology

Consider the Sierpinski space $$\text{S} = (\{0,1\}, \tau)$$ where $$\tau = \big\{\emptyset,\{0\}, \{0,1\}\big\}$$. Endow $$\text{S}^\omega$$ with the product topology.

If $$X, Y$$ are topological spaces with $$\text{S}^\omega \cong X \times Y$$, does this imply that there are $$\alpha, \beta \in \big(\omega\cup \{\omega\}\big)$$ such that $$X\cong \text{S}^\alpha$$ and $$Y \cong \text{S}^\beta$$?

(Note that $$\text{S}^0 = \text{S}^\emptyset$$ is the one-point space.)

• Is there an example where this fails for other topological spaces $S$? Jun 17, 2022 at 13:28
• No (to the OP's question): indeed if $K$ is a Cantor set and $D$ is the space consisting of a converging sequence along with its limit, then $K\simeq K\times D$ (actually $K\simeq K\times L$ for every nonempty closed subset of $K$).
– YCor
Jun 17, 2022 at 13:30

## 1 Answer

I think the answer is yes. I try to prove the stronger statement that every such homeomorphism is of the form $$S^{\mathbb{N}}=S^A\times S^B$$ for a disjoint union $$\mathbb{N}=A\amalg B$$.

Let me switch the roles of 0,1. This makes the following more easily readable.

We can introduce a order on $$S^w$$ by defining $$x\le y$$, iff each open set containing $$x$$ also contains $$y$$. We can do the same for $$X,Y$$ and note that the homeomorphism induces an order preserving map.

Furthermore $$S^w$$ is a lattice, e.g. any subset of it has a least upper bound and a greatest lower bound. If a product of two orders is a lattice, so are the two factors.

Especially $$S^w,X,Y$$ have a unique minimal element 0. Let $$f:X\times Y\rightarrow S^w$$ denote the homeomorphism. Write $$[a,b]$$ for $$\{c\mid a\le c\le b\}$$. For products of orders we have $$[(a_1,b_1),(a_2,b_2)]=[a_1,a_2]\times [b_1,b_2]$$. Thus if $$e_i$$ is the characteristic function of $$i\in \mathbb{N}$$, we have $$[0,e_i]$$ has only the two elements $$0,e_i$$. Pick $$(x,y)$$ with $$f(x,y)=e_i$$. We then have: $$[0,e_i]=[f(0,0),f(x,y)]=f([0,x]\times [0,y])$$ and hence exactly one of $$[0,x]$$ $$[0,y]$$ consists of two elements, and the other one of one element. If $$[0,x]$$ consists of two elements, put $$i$$ into $$A$$ otherwise, put it into $$B$$.

$$S^A=f(X\times 0)$$ follows by writing any element in $$S^A$$ as a least upper bound of $$e_i's$$ and using that $$f$$ is compatible with least upper bounds. Analogously $$S^B=f(0\times Y)$$ follows.

Last we have to show that for an arbitrary element $$g\in S^{\mathbb{N}}$$ we have $$g=f(x,y)$$, where $$f(x,0)=g_A,f(0,y)=g|_B$$ and $$g|_A$$ is the function which agrees with $$g$$ on $$A$$ and is zero otherwise. This again follows since $$f$$ is compatible with least upper bounds and the least upper bound of $$g|_A$$ and $$g|_B$$ is $$g$$, and the least upper bounds of $$(0,x)$$ and $$(0,y)$$ is $$(x,y)$$.