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 $A=f(F)$?



A link between continuity and 0-borelian?
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.


