Borel set preserved by continuous map 
Let $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$ be a continuous map. Show that if $A$ is a Borel subset of $\mathbb{R}^n$, then $f^{-1}(A)$ is a Borel subset of $\mathbb{R}^m$.

I know that for $A$ open subset of $\mathbb{R}^n$, then $f^{-1}(A)$ is open. Likewise for closed. A Borel subset is a countable unions/intersections of these sets, done repeatedly. From those facts, combined with $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$ and $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$, the statement should be intuitively true.
But to do it rigorously, I want to write $A$ as open and closed sets. But since the countable union/intersection can be done repeatedly, I don't know how I can write it that way.
 A: Another nice way of looking at this problem is to go through an "inductive" definition of the Borel subsets of $\mathbf{R}^n$. So, for ordinals $\alpha<\omega_1$ we inductively define the Borel hierarchy as follows:
Let $\mathbf{\Sigma}^0_1$ denote the collection of open subsets of $\mathbf{R}^n$.
Let $\mathbf{\Pi}^0_\alpha$ denote the collection of relative complements of elements of $\mathbf{\Sigma}^0_\alpha$.
Let $\mathbf{\Sigma}^0_\alpha$ denote the collection of $\bigcup_{n\in\mathbf{N}} X_n$ where each $X_n$ is in some $\mathbf{\Pi}^0_{\beta_n}$ with $\beta_n<\alpha$.
Recalling that the Borel subsets of $\mathbf{R}^n$ is smallest $\sigma$-algebra containing the open sets, we see that what we have done here is "stratify" the Borel sets. In particular, one can show that the Borel subsets of $\mathbf{R}^n$ is the same as the collection $\bigcup_{\alpha<\omega_1}\mathbf{\Sigma}^0_\alpha$. So now, we can in fact show that if $f:\mathbf{R}^m\rightarrow\mathbf{R}^n$ is continuous, then the levels of the Borel hierarchy are preserved under $f$. In particular, Borel sets are preserved by continuous preimages. Also note that, if we show $\mathbf{\Sigma}^0_\alpha$ sets are preserved under continuous preimages, then, since preimages are well behaved under complements, $\mathbf{\Pi}^0_\alpha$ sets are preserved under continuous preimages. So, it suffices to show that the $\mathbf{\Sigma}^0_\alpha$ sets are preserved under continuous preimages. We proceed by induction.
We know that the open sets, or $\mathbf{\Sigma}^0_1$ sets are preserved under continuous preimages, so assume that this is the case for all $\beta<\alpha$. Then, as we noted above, this holds for $\mathbf{\Pi}^0_\beta$ sets for all $\beta<\alpha$. So, let $A\in\mathbf{\Sigma}^0_\alpha$, then we see that $A=\bigcup_{n\in\mathbf{N}}X_n$ where each $X_n$ is in some $\mathbf{\Pi}^0_{\beta_n}$ where $\beta_n<\alpha$. Then, we see that $f^{-1}(A)=f^{-1}(\bigcup_{n\in\mathbf{N}}X_n)=\bigcup_{n\in\mathbf{N}}f^{-1}(X_n)$. But each $f^{-1}(X_n)$ is $\mathbf{\Pi}^0_{\beta_n}$ where $\beta_n<\alpha$, by our hypothesis. Thus, $f^{-1}(A)\in\mathbf{\Sigma}^0_\alpha$.
So, to answer your question (kind of), there is a nice way of expressing your original idea. 
A: I suggest a proof of your theorem in more general case.
Suppose $X$ and $Y$ are measurable and topological spaces respectively.
It is easy to prove fallowing lemma:
If $f$ map $X$ to $Y$, The Collection of all sets $E\subset Y$ such that $f^{-1}(E)$ is measurable set of $X$ is a $\sigma-$$Algebra$ of $Y$.
Now consider $\sigma-$$Algebra$ of $X$ is $Borel$ $\sigma-$$Algebra$ (ie suppose $X$ was a topological space in this case). Under this assumptions any continuous function $f$ is measurable.
Now Consider $Y$ equipped with It's $Borel$ $\sigma-$$Algebra$. According the above lemma and Continuity of $f$ such a $\sigma-$$Algebra$ of $Y$ (that defined in lemma) must contains $Borel$ $\sigma-$$Algebra$ of $Y$ (because of deffinition of $Borel$ $\sigma-$$Algebra$). So inverse image of any $Borel$ $subset$ of $Y$ is a $Borel$ $subset$ of $X$.
