# projections of analytic sets are analytic

## Projections along compact paved spaces

Given sets $X$ and $K$, the projection map $\pi_{X}\colon X\times K\rightarrow X$ is defined by $\pi_{X}(x,y)=x$. An important property of analytic sets  (http://planetmath.org/AnalyticSet2) is that they are stable under projections.

###### Theorem 1.

Let $(X,\mathcal{F})$ be a paved space (http://planetmath.org/PavedSpace), $(K,\mathcal{K})$ be a compact  (http://planetmath.org/PavedSpace) paved space and $\pi_{X}\colon X\times K\rightarrow X$ be the projection map.

If $S\subseteq X\times K$ is $\mathcal{F}\times\mathcal{K}$-analytic then $\pi_{X}(S)$ is $\mathcal{F}$-analytic.

The proof of this follows easily from the definition of analytic sets. First, there is a compact paved space $(K^{\prime},\mathcal{K}^{\prime})$ and a set $T\in\left(\mathcal{F}\times\mathcal{K}\times\mathcal{K}^{\prime}\right)_{% \sigma\delta}$ such that $S=\pi_{X\times K}(T)$. Then,

 $\pi_{X}(S)=\pi_{X}\left(\pi_{X\times K}(T)\right)=\pi_{X}(T).$

However, $(K\times K^{\prime},\mathcal{K}\times\mathcal{K}^{\prime})$ is a compact paved space (see products of compact pavings are compact (http://planetmath.org/ProductsOfCompactPavingsAreCompact)), which shows that $\pi_{X}(S)$ satisfies the definition of $\mathcal{F}$-analytic sets.

## Projections along Polish spaces

Theorem 1 above can be used to prove the following result for projections from the product    of a measurable space   and a Polish space  . For $\sigma$-algebras  (http://planetmath.org/SigmaAlgebra) $\mathcal{F}$ and $\mathcal{B}$, we use the notation $\mathcal{F}\otimes\mathcal{B}$ for the product $\sigma$-algebra (http://planetmath.org/ProductSigmaAlgebra), in order to distinguish it from the product paving $\mathcal{F}\times\mathcal{B}$.

###### Theorem 2.

Let $(X,\mathcal{F})$ be a measurable space and $Y$ be a Polish space with Borel $\sigma$-algebra (http://planetmath.org/BorelSigmaAlgebra) $\mathcal{B}$.

If $S\subseteq X\times Y$ is $\mathcal{F}\otimes\mathcal{B}$-analytic, then its projection onto $X$ is $\mathcal{F}$-analytic.

An immediate consequence of this is the measurable projection theorem.

Although Theorem 2 applies to arbitrary Polish spaces, it is enough to just consider the case where $Y$ is the space of real numbers $\mathbb{R}$ with the standard topology. Indeed, all Polish spaces are Borel isomorphic to either the real numbers or a discrete subset of the reals (see Polish spaces up to Borel isomorphism), so the general case follows from this.

If $Y=\mathbb{R}$, then the Borel $\sigma$-algebra is generated by the compact paving $\mathcal{K}$ of closed and bounded intervals. The collection  $a(\mathcal{F}\times\mathcal{K})$ of analytic sets is closed under  countable  unions and countable intersections  so, by the monotone class theorem, includes the product $\sigma$-algebra $\mathcal{F}\otimes\mathcal{B}$. Then, as the analytic sets define a closure operator   ,

 $a(\mathcal{F}\otimes\mathcal{B})\subseteq a(a(\mathcal{F}\times\mathcal{K}))=a% (\mathcal{F}\times\mathcal{K}).$

Thus every $\mathcal{F}\otimes\mathcal{B}$-analytic set is $\mathcal{F}\times\mathcal{K}$-analytic, and the result follows from Theorem 1.

Title projections of analytic sets are analytic ProjectionsOfAnalyticSetsAreAnalytic 2013-03-22 18:46:21 2013-03-22 18:46:21 gel (22282) gel (22282) 8 gel (22282) Theorem msc 28A05 MeasurableProjectionTheorem