Definition of analytic set in $\mathbb R^n$

Question

A standard definition of an analytic set in $\mathbb R^n$ is a set $A$ such that there exists a Polish space $Y$ and a Borel set $B\subseteq \mathbb R^n\times Y$, such that $A = \pi(B)$, where $\pi:\mathbb R^n\times Y\to\mathbb R^n$ is the natural projection $(x,y)\mapsto x$. Here of course $\mathbb R^n$ is endowed with the standard topology and $\mathbb R^n\times Y$ has the product topology.
Can I always pick $Y=\mathbb R$?
That is to say, for every analytic set $A$ in $\mathbb R^n$, is there a Borel set $B$ in $\mathbb R^{n+1}$ such that $A$ is the projection of $B$? Clearly, if such $B$ exists then $A$ is analytic, but is the other direction true? Or do I sometimes need a higher dimensional $Y$ to construct my Borel set?

Answer 1

Yes. Another standard definition of analytic set is one that is a projection of a Borel (we can even demand closed) subset of $\mathbb R^n\times \mathcal N,$ where $\mathcal N$ is Baire space. But we can homeomorphically embed $\mathcal N$ into $\mathbb R$ as the irrational numbers in $(0,1)$ so identify every Borel subset of $\mathbb R^n \times \mathcal N$ with a Borel subset of $\mathbb R^{n+1}$ with the same projection.

(Actually, we can say more. This shows every closed subset of $\mathbb R^n\times \mathcal N$ can be identified with a $G_\delta$ subset of $\mathbb R^{n+1}$, so by "we can even demand closed", "we can even demand $G_\delta$" in this case i.e. a subset of a Polish space $X$ is analytic if and only if it is a projection of a $G_\delta$ subset of $X\times \mathbb R$.)

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As to how to see this definition is equivalent, the hardest part I think is showing that any Borel set is a continuous image of $\mathcal N$, which can be obtained from the result that a Polish space is a continuous image of $\mathcal N$ by the trick of going to a finer topology where the set is closed (this is covered in Kechris). Then we have that any projection of a Borel set is a continuous image $f(\mathcal N)$ and we can consider that the graph of $f$ is a closed subset of $\mathcal N\times X.$ (This result is also covered in Jech lemma 11.6, but it looks like the proof given there is slightly different than what I had in mind.)