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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 $A=f(F)$?

Andrés E. Caicedo
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Dattier
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  • 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? – Pierre PC Nov 28 '19 at 19:23
  • yes, with my définition – Dattier Nov 28 '19 at 19:25
  • @Andrés E. Caicedo : if this result is already known could I have a reference? – Dattier Nov 29 '19 at 08:14

1 Answers1

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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.

Pierre PC
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  • Bravo........... – Dattier Nov 28 '19 at 19:36
  • what about the case, when you work with f(F), F a closed set ? – Dattier Nov 28 '19 at 19:39
  • I edited my answer accordingly. – Pierre PC Nov 28 '19 at 19:48
  • Brillant, it was already known ? – Dattier Nov 28 '19 at 19:51
  • I don't know, but it would very much surprise me if it wasn't. – Pierre PC Nov 28 '19 at 19:57
  • During my work, I discovred there exists some elementary ideas which is unknown or forgotten, for example : https://math.stackexchange.com/questions/2112322/a-new-characterization-of-an-annulus-in-the-plane https://math.stackexchange.com/questions/2114710/determine-the-general-term-of-a-recurrent-rational-sequence https://mathoverflow.net/questions/309019/does-every-real-function-have-this-weak-continuity-property – Dattier Nov 28 '19 at 20:01