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I'm confused by a subtle point in the definition of analytic sets. Suppose I have a Polish space $X$. Now I start with the collection of Borel sets in $X$ and take all their continuous images in $X$. Do I get the entire family of analytic sets in this way? In other words, can I say in good conscience that the analytic sets in $X$ are the continuous images of the Borel sets in $X$?

Let me state the question another way. By definition a set $A\subseteq X$ is analytic if it is the image $A=f(B)$ of some Borel set $B\subset P$ in some Polish space $P$ using some continuous mapping $f:P\to X$. I don't like referring to an external space $P$. What happens if I try to simplify the definition by requiring $P=X$; will I still get all the analytic sets?

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  • $\begingroup$ This is not really a simplification (i.e. is only a simplification in appearance), this is rather making the definition more rigid by making two different spaces play the same role. However the question makes sense and it might be interesting for its own sake. $\endgroup$
    – YCor
    Oct 20, 2021 at 10:20

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The answer is positive.

If $X$ is countable all subsets of $X$ are Borel, so they are their own continuous image through the identity function.

If $X$ is uncountable then it contains a copy of the Cantor space, but the Cantor space contains a copy of the Baire space $\mathcal N$ (which is necessarily $G_\delta$ in $X$, unrelated to your question but interesting nonetheless is the fact that a Polish space contains $\mathcal N$ as a closed subspace iff $X$ is not $\sigma$-compact) and being the continuous image of $\mathcal N$ is one of the many characterizations of analytic sets.

Edit: Following the comment by Samuel I realized that the definition used in the original question is different from the one I had in mind and the continuous function is required to have domain the whole of $X$. In that case the answer is negative: the Cook continuum is a compact metric space $X$ with the property that every continuous function $X\to X$ is either constant or the identity. Clearly no analytic but not Borel subset of $X$ is the image of a Borel set through a continuous function $X\to X$.

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  • $\begingroup$ Also unrelated to your question, but another way to characterize analytic sets without referring to a space external to $X$ is to say that $A\subseteq X$ is analytic iff $A$ can be obtained by applying Souslin's operation $\mathcal A$ to a Souslin scheme of closed subsets of $X$ $\endgroup$
    – Alessandro Codenotti
    Oct 20, 2021 at 15:59
  • $\begingroup$ But how do I know that a continuous function $f:\mathcal{N}\to X$ can be extended to a continuous function $f:X\to X$? Is this just some standard extension result for extending functions from zero-dimensional subspaces? $\endgroup$
    – Samuel
    Oct 21, 2021 at 21:33
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    $\begingroup$ @Samuel there is no such extension result, $\mathcal N$ embeds in $\Bbb R$ as the set of irrationals, but surely there are many continuous functions from the irrationals to the reals that don't extend. I didn't notice you required to have the whole of $X$ as domain and interpreted "continuous image of a Borel set" differently. If you insist on having the whole of $X$ as domain I believe that the answer is negative but I'm boarding a plane now and will update later with details! $\endgroup$
    – Alessandro Codenotti
    Oct 22, 2021 at 11:44
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    $\begingroup$ @Samuel plot twist, the plane was late and the answer is fixed $\endgroup$
    – Alessandro Codenotti
    Oct 22, 2021 at 12:04
  • $\begingroup$ Thanks, great answer! $\endgroup$
    – Samuel
    Oct 23, 2021 at 17:54

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