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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 $U \simeq \mathbb{Q}^\infty$.

Can anyone please help with this or provide any reading source? I would be really grateful.

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).

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

1 Answer 1

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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}<q_k<j_{n,k} \text{ for $k=1,\dots,n$}\}$, 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.

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  • $\begingroup$ 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 $\endgroup$ Jun 10 at 16:45
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    $\begingroup$ 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 $U$ $\endgroup$
    – Saúl RM
    Jun 10 at 21:42
  • $\begingroup$ 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/… $\endgroup$ Jun 11 at 18:59
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    $\begingroup$ 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 $\endgroup$
    – Saúl RM
    Jun 11 at 22:14

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