1

The canonical projection $\pi:\mathbb{R}^2\rightarrow \mathbb{R}$ such that $\pi(x,y)=x$

maps $G_\delta$ sets to Borel sets? i.e.

If $A=\cap_n^\infty A_n$ with $A_n$ open sets, then $\pi(A)$ is Borel?

William
  • 20,245
helmonio
  • 735
  • 1
    As William said, the continuous image of a Borel set need not be Borel. However, it is true that a continuous injective image of a Borel set (in particular a $G_\delta$ set) is Borel. – Francis Adams Aug 26 '14 at 19:41

1 Answers1

2

No. Even projections of $G_\delta$ sets could be non-Borel. A projection of a $G_\delta$ subset of $\mathbb{R} \times \mathbb{R}$ is called an analytic set. There are analytic non-Borel sets.

There are also many very concrete examples of analytic not Borel Sets. Often, they are more naturally subsets of other Polish spaces such as ${}^\omega \omega$; however, one can always transfer them into $\mathbb{R}$ by a Borel isomorphism.

For example, the set of reals that do not code well-ordering is an analytic not Borel set. Isomorphism of structures in the language with one binary relation symbol is also analytic. There are also numerous other examples from analysis concerning concepts like differentiability, summability, etc. See Classical Descriptive Set Theory Section 27 for other examples.

William
  • 20,245
  • thanks for answering! wanted to find out whether I can apply the result to prove that $\pi(f^{-1}(0,1))$ not is Borel measurable where $f=\limsup_{n\rightarrow \infty }f_n$ and $f_n: \mathbb{R}\rightarrow \mathbb{R}$ are continuous, although my teacher says that $\pi(f^{-1}(0,1))$ is Borel – helmonio Aug 26 '14 at 19:39
  • Projections of closed sets are always Borel, because every closed set is $\sigma$-compact, hence the same is true of its projection. – PhoemueX Aug 26 '14 at 20:04
  • @Helmonio: Do you mean $f_n : \Bbb{R}^2 \to \Bbb{R}$ instead of $f_n : \Bbb{R} \to \Bbb{R}$? – PhoemueX Aug 26 '14 at 20:08
  • @PhoemueX: Yes, I would greatly appreciate if you could guide me – helmonio Aug 26 '14 at 20:11
  • 1
    According to this post http://mathoverflow.net/questions/34142/projection-of-borel-set-from-r2-to-r1 , the analytic subsets of $\Bbb{R}$ are precisely the projections of $G_\delta$ subsets of $\Bbb{R}^2$. As there are non-Borel analytic sets, this shows that projections of $G_\delta$s are not Borel in general. – PhoemueX Aug 26 '14 at 20:21
  • @PhoemueX Yes. Analytic sets are projections of closed subsets of $\mathbb{R} \times {}^\omega \omega$ and not projection of closed subsets of $\mathbb{R} \times \mathbb{R}$. If the second coordinate space is compact (or $\sigma$-compact), the analytic sets are generally the projection of $G_\delta$ sets. I will edit the post shortly. – William Aug 26 '14 at 20:23
  • @PhoemueX: “the set of reals that do not code well-ordering”? What does this mean? What does this set look like? – Bear and bunny Sep 03 '15 at 21:25