Continuous image of $\mathbb{R}$ is Borel.

Question

Let $X$ be a Polish space and $f\colon\mathbb{R}\to X$. Then since $\mathbb{R}=\bigcup_{n=-\infty }^\infty [n,n+1]$ - a countable union of compact set, $f(X)$ is a countable union of compact sets $\Rightarrow$ $f(X)$ is $F_\sigma$$\Rightarrow$ Borel.

I think that there is a problem with this argument as it can be easily extended to open sets in $\mathbb{R}^n$ and their images are not always Borel. I need help.

Many thanks in advance!

Of course, thank you. I first meant to ask the question about the Baire space but then reconsidered. — peter, Apr 27 '14 at 17:39

score 4 · Answer 1 · answered Apr 27 '14 at 13:02

4

The argument is correct, and it extends to $\Bbb R^n$ as well, and to every other $\sigma$-compact Polish space.

The "problem" that you point to is that not every Borel set is $\sigma$-compact, so this is not true in general for every Borel subset of $\Bbb R^n$, and not every Polish space is $\sigma$-compact (e.g. $\omega^\omega$ is not $\sigma$-compact) so it's not true in general for every Polish space.

But there's no real issue with $\Bbb R^n$.

answered Apr 27 '14 at 13:02

Asaf Karagila

393,674

Thanks! I probably misunderstood Michael Greinecker's answer to Continuous images of open sets are Borel? – peter Apr 27 '14 at 17:43
Yes, you have. In that example he takes a Borel set whose projection is not Borel, and changes the topology to make it open. The new topology is not the standard topology on $\Bbb R^n$. – Asaf Karagila Apr 27 '14 at 17:46

Continuous image of $\mathbb{R}$ is Borel.

1 Answers1