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.)

  • 1
    $\begingroup$ Is there an example where this fails for other topological spaces $S$? $\endgroup$ Jun 17, 2022 at 13:28
  • 1
    $\begingroup$ 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 $K$). $\endgroup$
    – YCor
    Jun 17, 2022 at 13:30

1 Answer 1


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 $(x,y)$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.