# Quotients of the irrationals

Everyone knows that there is a closed equivalence relation $$\sim$$ on the Cantor set $$C$$ such that each non-trivial equivalence class has exactly $$2$$ points and $$[0,1]\simeq C/\sim$$. Thus a closed quotient of a zero-dimensional space may not be zero-dimensional, even if the equivalence classes are compact. Note that in this case, if we specify that $$C/\sim$$ must be totally separated, then $$C/\sim$$ is automatically zero-dimensional, because every totally separated compact metric space is zero-dimensional. Totally separated means that for every two points $$x$$ and $$y$$ in the space, there is a clopen set containing $$x$$ and missing $$y$$. Zero-dimensional means that the space has a basis of clopen sets.

This question is about similar quotients of the irrationals $$\mathbb P$$.

Question. Let $$\sim$$ be a closed equivalence relation on $$\mathbb P$$ such that $$\mathbb P/\sim$$ is Polish and every equivalence class is compact. If $$\mathbb P/\sim$$ is totally separated, then is $$\mathbb P/\sim$$ necessarily zero-dimensional?

Note that every Polish space is a closed quotient of $$\mathbb P$$; shown here. So the condition that the equivalence classes are compact is critical. I believe my question is equivalent to: Is every totally disconnected Polish perfect image of $$\mathbb P$$ zero-dimensional? A continuous mapping is perfect if it is closed has compact point preimages.

• "Everyone knows": is it an exercise? the obvious relation I can think of has countably many equivalence classes of cardinal 2 and others have cardinal 1. – YCor Apr 2 '20 at 21:42
• @YCor that's what I meant by non-trivial equivalence classes have size 2 – D.S. Lipham Apr 2 '20 at 21:43
• @WlodAA Engelking's definition (which I use) is that the natural mapping $X\to X/\sim$ is closed. This guarantees $X/\sim$ is metrizable if $X$ is metrizable. I think he gives an example showing this definition is stronger than yours. – D.S. Lipham Apr 3 '20 at 2:35
• I am justified in not knowing -- I am the only Polish topologist who at the time was not gifted their own copy from Engelking of his classic. He told me that he wanted to give me the new edition but... then crazy political times came and I emigrated and he had more urgent matters to attend to. In this and only this way I am a unique Polish topologist. – Wlod AA Apr 3 '20 at 2:42
• @WlodAA Your comment about Polish spaces is correct. Engelking proved that every Polish space is a continuous closed image of $\mathbb P$, and hence a closed quotient of $\mathbb P$. I've simplified my question now. – D.S. Lipham Apr 3 '20 at 2:53

Theorem. A topological space $$X$$ is an image of the space of irrationals $$\mathbb P$$ under a perfect map $$f:\mathbb P\to X$$ if and only if $$X$$ is Polish and nowhere locally compact.
Proof. To prove the "only if" part, assume that a topological space $$X$$ is the image of $$\mathbb P$$ under a perfect map $$f:\mathbb P\to X$$. By Theorem 3.7.20 of Engelking's "General Topology" (denoted later by [EGT]), the space $$X$$ is regular, by Theorem 3.7.19 in [EGT], $$X$$ is second-countable and by the Uryssohn Metrization Theorem, $$X$$ is metrizable and separable. By Theorem 3.9.10 of [EGT], the space $$X$$ is Cech-complete and being separable and metrizable is Polish. Assuming that $$X$$ contains a compact subset $$K$$ with non-empty interior, we can apply Theorem 3.7.2 of [EGT] and conclude that the preimage $$f^{-1}(K)$$ is compact and by the continuity of $$f$$, it has non-empty interior in $$\mathbb P$$. On the other hand, it is well-known that $$\mathbb P$$ contains no compact sets with nonempty interior. This contradiction shows that $$X$$ is nowhere locally compact.
To prove the "if" part, assume that $$X$$ is a nowhere locally compact Polish space. Let $$\bar X$$ be any metric compactification of $$X$$. By Theorem 4.18 in Kechris' "Classical Descriptive Set Theory" (denoted later by [CDST]), $$\bar X$$ is the image of the Cantor cube under a continuous map $$\tilde g:2^\omega\to\bar X$$. Using the Kuratowski-Zorn Lemma, choose a minimal closed subset $$Z\subseteq 2^\omega$$ such that $$f(Z)=\bar X$$. Let $$g=\tilde g{\restriction}Z$$. The minimality of $$Z$$ and closedness of the map $$g:Z\to\bar X$$ ensure that for any non-empty open set $$U\subset Z$$ the image $$g(U)$$ has nonempty interior in $$\bar X$$.
Being Polish, the space $$X$$ is a $$G_\delta$$-set in $$\bar X$$ and its preimage $$P=g^{-1}(X)$$ is a $$G_\delta$$-set in the zero-dimensional compact space $$Z$$. We claim that $$P$$ is nowhere locally compact. Assuming that $$P$$ contains a compact subset $$K$$ with nonempty interior, we can use the minimality of $$Z$$ and conclude that $$g(K)$$ is a compact set with nonempty interior in $$X$$, which contradicts the nowhere local compactness of $$X$$. This contradiction shows that the Polish zero-dimensional space $$P$$ is nowhere locally compact. By the Aleksandrov-Urysohn characterization of $$\mathbb P$$ (see Theorem 7.7 in [CDST]), the space $$P$$ is homeomorphic to the space of irraionals $$\mathbb P$$. It remains to apply Proposition 3.7.6 of [EGT] to see that the restriction $$f=g\restriction P:P\to X$$ is perfect.