# projections of analytic sets are analytic

## Projections along compact paved spaces

Given sets $X$ and $K$, the projection map ${\pi}_{X}:X\times K\to X$ 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 $\mathrm{(}X\mathrm{,}\mathrm{F}\mathrm{)}$ be a paved space (http://planetmath.org/PavedSpace), $\mathrm{(}K\mathrm{,}\mathrm{K}\mathrm{)}$ be a compact^{} (http://planetmath.org/PavedSpace) paved space and ${\pi}_{X}\mathrm{:}X\mathrm{\times}K\mathrm{\to}X$ be the projection map.

If $S\mathrm{\subseteq}X\mathrm{\times}K$ is $\mathrm{F}\mathrm{\times}\mathrm{K}$-analytic then ${\pi}_{X}\mathit{}\mathrm{(}S\mathrm{)}$ is $\mathrm{F}$-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 $\mathrm{(}X\mathrm{,}\mathrm{F}\mathrm{)}$ be a measurable space and $Y$ be a Polish space with Borel $\sigma $-algebra (http://planetmath.org/BorelSigmaAlgebra) $\mathrm{B}$.

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

Canonical name | ProjectionsOfAnalyticSetsAreAnalytic |

Date of creation | 2013-03-22 18:46:21 |

Last modified on | 2013-03-22 18:46:21 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 8 |

Author | gel (22282) |

Entry type | Theorem |

Classification | msc 28A05 |

Related topic | MeasurableProjectionTheorem |