A topological space $X$ is called analytic if it is a continuous image of a Polish space, i.e., the image of a Polish space $P$ under a continuous surjective map $f:P\to X$.

We say that a topological space $X$ is a Borel image of a Polish space if $f(P)=X$ for some Borel surjective function $f:P\to X$ defined on a Polish space $P$.

We recall that a function $f:X\to Y$ between topological spaces is Borel if the preimage $f^{-1}(U)$ of any open set $U\subset Y$ is Borel in $X$.

Problem 1. Is a (regular) topological space $X$ analytic if it is a Borel image of a Polish space $P$? Will be the answer affirmative if $X$ is a (Baire) topological group?

The following theorem reduces the problem to finding a countable network for the space $X$.

Let us recall that a family $\mathcal N$ of subsets of a topological space $X$ is a network for $X$ if for any open set $U\subset X$ and point $x\in U$ tehre exists a set $N\in\mathcal N$ such that $x\in N\subset U$.

Theorem. A regular topological space $X$ is analytic if and only if $X$ has a countable network and $X$ is a Borel image of a Polish space.

Proof. Let $\mathcal N$ be a countable network for $X$. Since $X$ is regular, we can replace each set $N\in\mathcal N$ by its closure $\bar N$ and assume that $\mathcal N$ consists of closed subsets of $X$.

Let $f:P\to X$ be a Borel surjective map defined ona Polish space. It follows that $\{f^{-1}(N):N\in\mathcal N\}$ is a countable family of Borel subsets of $P$. By Exercise 13.5 in the "Classical Descriptive Set Theory" of Kechris, there exists a bijective continuous map $g:Z\to P$ from a Polish space $Z$ such that each set $f^{-1}(N)$, $N\in\mathcal N$, is clopen. Then the map $f\circ g:Z\to X$ is continuous, witnessing that $X$ is an analytic space. $\square$

Therefore, Theorem reduces Problem 1 to the following equivalent

Problem 2. Assume that a regular space $X$ is Borel image of a Polish space. Has $X$ a countable retwork?

I can only prove (using the Four Poles Theorem) that $X$ has countable spread (which means that $X$ each discrete subspace of $X$ is countable).

Probem 3. Assume that a (regular) topological space $X$ is a Borel image of a Polish space.

1) Is $X$ hereditarily Lindelöf?

2) Is $X$ hereditarily separable?

3) Is $X\times X$ a Borel image of a Polish space?

Added in Edit. Under PFA the answer to Problem 3(1) is affirmative (because PFA implies that each regular space with countable spread is hereditarily Lindelof, see Theorem 8.10 in Todorcevic's book "Partition Problems in Topology").


1 Answer 1


I have found a simple counterexample to Problems 1 and 2 (maybe it will be helpful to other researchers):

Fact. The Sorgenfrey line $\mathbb S$ does not have countable network (and hence is not analytic), but it is the image of the real line under the identity map $\mathbb R\to\mathbb S$, which is Borel (because of hereditary Lindeloffness of $\mathbb S$).

Observe that the Sorgenfrey line is a Baire paratopological group. What about Baire topological groups?

Problem. Is a Baire topological group $X$ Polish if $X$ is a Borel image of a Polish space?

Remark. By an old result of Christensen, a Baire topological group $X$ is Polish if and only if $X$ is a continuous image of a Polish space. The following theorem gives an affirmative answer to the Problem for topological groups of countable pseudocharacter.

Theorem 1. A topological group $X$ is Polish if and only if it is Baire, has countable pseudocharacter and is a Borel image of a Polish space.

Using a result of Todorcevic that under PFA each Hausdorff space of countable spread has a countable pseudocharacter, it is possible to improve the above theorem and prove

Theorem 1'. Under PFA a Baire topological group $X$ is Polish if and only if it is a Borel image of a Polish space.

So, Problem has an affirmative answer under PFA and the problem is to find an answer in ZFC (or a counterexample under CH).

  • 1
    $\begingroup$ Sorry, it's still early, in my brain I've read "the long line" instead of "the Sorgenfrey line". Still, though. Very interesting, since it would seem reasonable that a Borel image of a Polish space will be analytic. But I guess that this happens only when you can make the Borel function continuous by refining the topology while remaining Polish. So I guess this is not the case in general, which is interesting. $\endgroup$
    – Asaf Karagila
    Jan 12, 2019 at 8:37
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    $\begingroup$ About the PFA situation, it sort of reminds me of the choiceless results on automatic continuity between Polish groups, which itself is a theorem of ZF+DC: a Baire measurable or a Borel measurable group homomorphism between Polish groups is continuous. In fact the target group need only be normed, not complete. $\endgroup$
    – Asaf Karagila
    Jan 12, 2019 at 8:42
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    $\begingroup$ @AsafKaragila I hope that such pathologies (like those with Sorgenfrey line) cannot appear for topological groups (which is my main motivation), but at the moment I can proof only consistent PFA-results, but not in ZFC. It is interesting if these "Borel" questions can lead to independence results. Usually, evering related to Borel sets is absolute... $\endgroup$ Jan 12, 2019 at 8:44
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    $\begingroup$ @AsafKaragila In fact, my aim is to prove this PFA-characterization of Polish groups (as Baire topological groups which are Borel images of Polish spaces) in ZFC. Maybe try to prove that this statement is absolute? $\endgroup$ Jan 12, 2019 at 8:48
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    $\begingroup$ I'm not sure how absoluteness would help, since forcing PFA requires you to start with some fairly large cardinals. Moreover, since a Polish group does not have size greater than the continuum, I'm not sure how PFA does so much more than MA. So that would be my first objective, I guess. $\endgroup$
    – Asaf Karagila
    Jan 12, 2019 at 8:50

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