Continuous functions and Borel sets 
*

*Is every Borel set a preimage of some Borel set under some continuous and bounded function?
That is, for a given topological space $X$ does it hold, that $$\forall_{\ B \ Borel \subset X} \ \exists_{\ C \ Borel \subset \mathbb{R}^n, \ \ f \in C_0(X, \ \mathbb{R}^n)} \ \  B=f^{-1}(C)$$?

*Is Borel $\sigma$-field on $X$ the smallest $\sigma$-field that makes all continuous, compactly supported functions $\in C_c(X, \ \mathbb{R}^n)$ measurable (given the usual Borel $\sigma$-algebra on $\mathbb{R}^n$)?

If it's necessary, you can assume $X = \mathbb{R}^m$ or a metric space (my main interests), but I'm also interested in more genral cases.
 A: I'll prove these for $X=\mathbb R^m$, $m=n$.

*

*Yes. Let $f:X \to (0, 1)^m$ be continuous with continuous inverse; we can take e.g.
$$f(x)_i = 1/(1+e^{-x_i}).$$ Then for any Borel set $B \subseteq X$, the image $C = f(B) = (f^{-1})^{-1}(B)$ is Borel, and $B = f^{-1}(C)$.


*Yes. Let $A_n = [-n, n]^m$, and choose continuous, compactly supported $f_n:X \to \mathbb R^m$ such that the image $I_n = f_n(A_n)$ is Borel, and the restriction $f_n\mid_{A_n}:A_n \to I_n$ has continuous inverse; we can take e.g.
$$f_n(x) = f(x)g_n(x), \qquad g_n(x) = \max(0, \min(1, n + 1 - \max( \lvert x_i \rvert))).$$
Then for any Borel set $B \subseteq A_n$, we have $B = g_n^{-1}(\{1\}) \cap f_n^{-1}(f_n(B))$, so $B \in \sigma(C_c(X, \mathbb R^m))$. We deduce this $\sigma$-algebra contains all bounded Borel sets, and so by countable union all Borel sets.
Edit: for $m < n$, the same proofs hold, e.g. by setting the extra coordinates of $f$ to zero. For $m > n$, result 2. holds as in Jkbb's answer. For result 1., in the case $m=2$, $n=1$ the answer is no.
Let $D \subset \mathbb R^2$ be the open unit disc, and define the Borel set $B = D \cup \{(1, 0)\}$. Suppose we have continuous $f$ and Borel $C$ satisfying $f^{-1}(C)=B$.
By translation and scaling, we may assume $f(1, 0) = 0$, and $f > 0$ somewhere in $\partial B$. Applying the intermediate value theorem to a path along $\partial B$, we deduce that $0 \in C$, and $C \cap (0,\epsilon) = \emptyset$ for some $\epsilon > 0$. We thus have that $f \le 0$ on $B$, and $f > 0$ on $\partial B \setminus B$, contradicting the continuity of $f$.
A: Here is the answer to the 2. point (it's affirmative): lets take any bounded open ball $B$ in $\mathbb{R}^n$ and define $f:\mathbb{R}^n \rightarrow \mathbb{R}$, $f(x) = dist(x, \ \partial B)$ for $x \in  B$ and $f(x) = 0$ for $x \in \mathbb{R}^n / B$.  It's continuous and compactly supported.
Now lets define $F: \mathbb{R}^n \rightarrow \mathbb{R}^m$ componentwise so that $F_i = f$ for $i \in \{1, 2, ..., m\}$. Then $F^{-1}((0, \infty]^m) = B$.
Now the smallest $\sigma$-algebra containing all the bounded open balls is the Borel $\sigma$-algebra (since $\mathbb{R}^n$ is Lindelof and we have the countable union property of $\sigma$-algebras).
I would still like to know the answer to the 1. question though.
