# Why can any open subset $U$ of $\mathbb{Q}^\infty$ be written as disjoint union of basic clopen subsets?

I am reading Engelen´s paper and have trouble with this proof of Lemma 2.1 (a) (link is below).

It is easily seen that any non-empty open subspace $$U$$ of $$\mathbb{Q}^\infty$$ can be written as an infinite disjoint union of non-empty basic clopen subsets; hence, $$U = \mathbb{N} \times \mathbb{Q}^\infty \simeq \mathbb{Q}^\infty$$.

In particular, I don't see how the $$U$$ can be written as the said disjoint union and how that implies that $$U \simeq \mathbb{Q}^\infty$$.

Definition: $$\mathbb{Q}^\infty$$ is defined as a set of all rational sequences, endowed with the standard product topology.

Source: Engelen - Countable Product of zero-dimensional absolute $$F_{\sigma \delta}$$ spaces, Lemma 2.1 (a).

• "endowed with the standard product topology": but product of which topology on $\mathbf{Q}$? the discrete one or the one induced by inclusion in $\mathbf{R}$?
– YCor
Jun 9 at 17:01
• @YCor The one "inherited" from $\mathbb{R}$. Jun 9 at 17:02

We can first express the open set $$U$$ as a countable union of basic clopen subsets $$A_n=\{(q_k)_{k\in\mathbb{N}};i_{n,k}, where $$i_{n,k}$$ and $$j_{n,k}$$ are irrational numbers. Some $$A_n$$ may be empty.

To answer the question it is enough to express $$A_n\setminus\bigcup_{i=1}^{n-1}A_i$$ as a finite union of disjoint basic clopens. To do it, we can work in $$\mathbb{Q}^n$$ instead of $$\mathbb{Q}^\infty$$ due to the construction of the sets $$A_n$$.

Writing $$\mathbb{Q}^n=\mathbb{Q}_1\times\dots\times\mathbb{Q}_n$$, notice that for each $$k\leq n$$, the irrationals $$i_{1,k},\dots,i_{n,k},j_{1,k},\dots,j_{n,k}$$ give a partition of $$\mathbb{Q}_k$$ into finitely many clopen subsets. Taking a product of these partitions, we obtain a partition of $$\mathbb{Q}^n$$ into disjoint clopens such that any of the $$A_i$$, with $$i\leq n$$, is a finite union of these clopens. So $$A_n\setminus\bigcup_{i=1}^{n-1}A_i$$ is also a finite union of the clopens.

If we pick the $$A_n$$ so that no finite union of them covers $$U$$, we also get infinitely many non empty clopens.

This implies that $$U\equiv\mathbb{Q}^\infty$$ because any basic clopen set like the ones above is homeomorphic to $$\mathbb{Q}^\infty$$, which can be deduced using that any clopen subset of $$\mathbb{Q}$$ is homeomorphic to $$\mathbb{Q}$$ by Sierpinski's theorem.

• Thank you, this is wonderful! The only part I dont understand is how the "This implies that $U\equiv\mathbb{Q}^\infty$..." follows from that we get infinitely many clopens if we pick $A_n$ so that no finite union covers $U$? How do we "get infinite many non empty clopens"? Thank you Jun 10 at 16:45
• The procedure always gives $U$ as a union of clopens, the problem is that only finitely many of them may be nonempty (for example this happens if $A_1=U$). However this cannot happen if $U$ is not contained in any finite union of the $A_n$, because any finite union of the clopens we obtain is contained in a finite union of the sets $A_n$, so it can't be the whole $U$ Jun 10 at 21:42
• Thank you! I am still quite unsure, after consulting that with colleagues, so I tried to rephrase your answer to the initial question on stackexchange and gave you credits. I am not sure if you have account there, but I highlighted what is still unclear and will appreaciate if you take a look. math.stackexchange.com/questions/4468793/… Jun 11 at 18:59
• Yeah this proof was written a bit too fast as I was in the middle of an exam week. I have tried to write it in more detail in MSE Jun 11 at 22:14