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Is it true that :

1/ if $f$ real continuous and $O$ an open set then $f(O)$ is a 0-borelian?

2/ if $A$ a 0-borelian set then there exists $f$ real continuous and $O$ an open set with $A=f(O)$?

$B$ a 0-borelian of $\mathbb R$, have the form $B= \bigcup \limits_{n \in \mathbb N} F_n$ with $F_n$ closed set.

what about the case :

1/ if $f$ real continuous and $F$ a closed set set then $f(F)$ is a 0-borelian?

2/ if $A$ a 0-borelian set then there exists $f$ real continuous and $F$ a closed set with $A=f(F)$?

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  • $\begingroup$ Your claim about the form of Borelian subsets is not true, for instance the set of rationals is a counterexample. Are you interested in the question with the sets you describe, or actual Borelian subsets? $\endgroup$
    – Pierre PC
    Nov 28, 2019 at 19:23
  • $\begingroup$ yes, with my définition $\endgroup$
    – Dattier
    Nov 28, 2019 at 19:25
  • $\begingroup$ @Andrés E. Caicedo : if this result is already known could I have a reference? $\endgroup$
    – Dattier
    Nov 29, 2019 at 8:14

1 Answer 1

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1) Yes, because every open set in $\mathbb R$ is a countable union of compact sets, whose image is compact (hence closed).

2) No, because the usual Cantor set $C$ (uncountable, totally disconnected) is closed, and any continuous function $f:O\to C$ must be constant on each of the (countably many) connected components of the open set $O$.


Replacing the open sets at the source by closed sets, then

1) still holds, because every closed set is a countable union of compact subsets.

2) becomes true, because every countable union of closed sets is a countable union of closed bounded sets, say $(K_1,K_2,\ldots)$. Then each $K_i$ is the image of some affine function defined on a closed subset of $(i-1/2,i+1/2)$ (just fit a rescaled version of $K_i$ into this interval), so patching all these functions and closed sets together leads the result.

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  • $\begingroup$ Bravo........... $\endgroup$
    – Dattier
    Nov 28, 2019 at 19:36
  • $\begingroup$ what about the case, when you work with f(F), F a closed set ? $\endgroup$
    – Dattier
    Nov 28, 2019 at 19:39
  • $\begingroup$ I edited my answer accordingly. $\endgroup$
    – Pierre PC
    Nov 28, 2019 at 19:48
  • $\begingroup$ Brillant, it was already known ? $\endgroup$
    – Dattier
    Nov 28, 2019 at 19:51
  • $\begingroup$ I don't know, but it would very much surprise me if it wasn't. $\endgroup$
    – Pierre PC
    Nov 28, 2019 at 19:57

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