# proof of countable unions and intersections of analytic sets are analytic

Let $(X,\mathcal{F})$ be a paved space and ${({A}_{n})}_{n\in \mathbb{N}}$ be a sequence^{} of $\mathcal{F}$-analytic sets^{} (http://planetmath.org/AnalyticSet2). We show that their union and intersection^{} is also analytic.

From the definition of $\mathcal{F}$-analytic sets, there exist compact paved spaces $({K}_{n},{\mathcal{K}}_{n})$ and ${S}_{n}\in {(\mathcal{F}\times {\mathcal{K}}_{n})}_{\sigma \delta}$ such that

$${A}_{n}=\{x\in X:(x,y)\in {S}_{n}\text{for some}y\in {K}_{n}\}.$$ |

We start by showing that ${\bigcap}_{n}{A}_{n}$ is analytic. Let $K={\prod}_{n}{K}_{n}$ and $\mathcal{K}={\prod}_{n}{\mathcal{K}}_{n}$ be the product paving, and ${\pi}_{n}:K\to {K}_{n}$ be the projection map. Then $x\in {\bigcap}_{n}{A}_{n}$ if and only if for each $n$ there is a ${y}_{n}\in {K}_{n}$ with $(x,{y}_{n})\in {S}_{n}$. Equivalently, setting $y=({y}_{1},{y}_{2},\mathrm{\cdots})$, then $(x,y)\in {\bigcap}_{n}{\pi}_{n}^{-1}({S}_{n})$. However, this is in ${(\mathcal{F}\times {\mathcal{K}}_{n})}_{\sigma \delta}$ and we can write,

$$\bigcap _{n}{A}_{n}={\pi}_{X}\left(\bigcap _{n}{\pi}_{n}^{-1}({S}_{n})\right),$$ |

where ${\pi}_{X}:X\times K\to X$ is the projection map. As products of compact pavings are compact, $(K,\mathcal{K})$ is compact and it follows from the definition that ${\bigcap}_{n}{A}_{n}$ is $\mathcal{F}$-analytic.

We now show that ${\bigcup}_{n}{A}_{n}$ is analytic. Let $K={\sum}_{n}{K}_{n}$ and $\mathcal{K}={\sum}_{n}{\mathcal{K}}_{n}$ be the direct sum^{} paving, which is compact (http://planetmath.org/SumsOfCompactPavingsAreCompact). Also, write ${S}_{n}={\bigcap}_{m=1}^{\mathrm{\infty}}{T}_{m,n}$ for ${T}_{m,n}\in {(\mathcal{F}\times {\mathcal{K}}_{n})}_{\sigma}$.
We identify ${K}_{n}$ with a subset of $K$, so that $K$ is the union of the disjoint sets ${K}_{n}$.
Then $x\in {\bigcup}_{n}{A}_{n}$ if and only if $(x,y)\in {S}_{n}$ for some $n$ and some $y\in K$,

$$\bigcup _{n}{A}_{n}={\pi}_{X}\left(\bigcup _{n}{S}_{n}\right).$$ |

However, the fact that ${K}_{{n}_{1}},{K}_{{n}_{2}}$ are disjoint for ${n}_{1}\ne {n}_{2}$ says that ${T}_{m,{n}_{1}},{T}_{m,{n}_{2}}$ are disjoint and, therefore,

$$\bigcup _{n}{S}_{n}=\bigcup _{n}\bigcap _{m}{T}_{m,n}=\bigcap _{m}\bigcup _{n}{T}_{m,n}\in {\left(\mathcal{F}\times \mathcal{K}\right)}_{\sigma \delta}.$$ |

So ${\bigcup}_{n}{A}_{n}$ is $\mathcal{F}$-analytic.

Title | proof of countable unions and intersections of analytic sets are analytic |
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Canonical name | ProofOfCountableUnionsAndIntersectionsOfAnalyticSetsAreAnalytic |

Date of creation | 2013-03-22 18:46:19 |

Last modified on | 2013-03-22 18:46:19 |

Owner | gel (22282) |

Last modified by | gel (22282) |

Numerical id | 4 |

Author | gel (22282) |

Entry type | Proof |

Classification | msc 28A05 |