If $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous, 1-1, and on-to, then f maps Borel Sets to Borel Sets Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be continuous, on-to, and 1-1. Prove that if $A$ is a Borel set, then $f(A)$ is a Borel set.
 A: Let's not worry about whether $f$ is a homeomorphism, it ends up not playing a role. 
First, $f(C)$ is Borel when $C$ is closed. This is clear if $C$ is compact, because continuous maps send compact sets to compact (and therefore closed) sets. If $C$ is unbounded, we cannot quite say that $f(C)$ is closed by the same argument, but note that $C=\bigcup_n(C\cap[-n,n])$, so $f(C)=\bigcup_n f(C\cap[-n,n])$ is a countable union of closed sets.
It follows that $f(O)$ is Borel is $O$ is open, because $f(O)=f(\mathbb R)\setminus f(\mathbb R\setminus O)$ is a difference of Borel sets (since both $\mathbb R$ and $\mathbb R\setminus O$ are closed). Note that we do not even need to use the assumption that $f$ is onto, but we are definitely using that $f$ is injective.
Let $\mathcal F$ be the collection of Borel sets $A$ such that $f(A)$ is also Borel. We have shown that $\mathcal F$ contains the open sets. 
Note now that $\mathcal F$ is closed under complements. This is because if $f(A)$ is Borel, then so is $f(\mathbb R\setminus A)=f(\mathbb R)\setminus f(A)$ (again, the equality holds because $f$ is injective). 
Finally, $\mathcal F$ is closed under countable unions, because if $f(A_n)$ is Borel for each $n$, then so is $f(\bigcup_n A_n)=\bigcup_n f(A_n)$.
This means that $\mathcal F$ is the $\sigma$-algebra of all Borel sets of reals, that is, $f(A)$ is Borel for each Borel set $A$.

Hmm... It turns out the question is a duplicate so it may end up being closed. Let me add some remarks just in case: If you want to argue that $f$ is a homeomorphism, which would certainly simplify things, note that since $f$ is injective, it must be monotone, and therefore (by continuity) it maps open intervals to open intervals. It follows that $f^{-1}$ is continuous.
This simplifies the argument, because we get not just that $f(A)$ is Borel if $A$ is Borel, but in fact we see that $f(A)$ has the same Borel complexity as $A$: If $A$ is open, so is $f(A)$, if $B$ is $F_\sigma$, so is $f(B)$, etc.
Trevor hints in a comment that the above is anyway not the most direct approach. As you can see in the answers to the other question (the one this duplicates), descriptive set theory provides a very elegant alternative approach, via a theorem of Suslin, that if both a set $A$ and its complement are continuous images of a Borel set, then in fact $A$ is Borel. 
A: Hint: Let $C$ be the collection of subsets $A \subset \mathbb{R}$ such that $f[A]$ is Borel.  Show that $C$ contains all the open sets and is closed under complementation and countable unions.
