Proof that a homeomorphic image of a non-borel set is non-borel This question seeks to expand the proof given in the answer to this question.
I am weak in topology, and am wondering if someone can provide a proof of why a homeomorphic image of a non-borel set is non-borel?
 A: Since a homeomorphism $f: X \to Y$ is continuous, it is also Borel-measurable. The same is true of $f^{-1}: Y \to X$.  
Let $\mathcal{B}_{x}$ and $\mathcal{B}_{y}$ be the Borel $\sigma$-algebras on $X$ and $Y$ respectively. 
Since $f$ is measurable we know that for every $B_{1} \in \mathcal{B}_{y}$ we have that $f^{-1}(B_{1}) \in \mathcal{B}_{x}$, and that for every $B_{2} \in \mathcal{B}_{x}$ we have that $f(B_{2}) \in \mathcal{B}_{y}$. 
Let $B_{0} \subset X$ be a non-Borel set. Then $B_{0} \notin \mathcal{B}_{x}$. Since for every $B_{1} \in \mathcal{B}_{y}$ we have that $f^{-1}(B_{1}) \in \mathcal{B}_{x}$, it must be that there is no set in $\mathcal{B}_{y}$ that is mapped to $B_{0}$ (i.e. no Borel set in $Y$ can be mapped by $f$ to a non-Borel set in $X$). 
The same holds if $B_{0} \subset Y$ is not Borel; no Borel set in $X$ can be mapped by $f$ to a non-Borel set in $Y$. 
By the fact that $f$ is bijective, this implies that no non-Borel set $B_{0} \subset X$ can be mapped to a Borel set $f(B_{0})$ as otherwise $f^{-1} (f (B_{0}))= B_{0}$ must be Borel (since borel sets are mapped to borel sets), which is a contradiction. The same holds for non-borel sets $B_{0} \subset Y$. 
Therefore, a homeomorphic image of a non-borel set must be non-borel.
