The Eigenspace

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/Part 3: Measure Theory/Chapter 4: Set Systems/Section 4.3: Borel $\s$-Algebra

Lemma 4.3.2.label Let $X$ be a topological space, then $\cali{B}_{X}$ is the smallest subset of $2^{X}$ that:

  1. (1)

    contains open and closed subsets of $X$.

  2. (2)

    is closed under countable intersecttions.

  3. (3)

    is closed under countable disjoint unions.

Proof. Let $\cali{F}\suf 2^{X}$ be the smallest subset of $2^{X}$ satisfying (1,2,3) then $\cali{F}\suf\cali{B}_{X}$. OTOH, let

\begin{align*}\cali{F}_{0}=\curl{A\suf X:A\in\cali{F},A^c\in\cali{F}}\suf\cali{F}\end{align*}

$\emp,X\in \cali{F}_{0}$ since $\cali{F}$ satisfies (1). By definition $\cali{F}_{0}$ is closed under complements. Let $\curl{A_n }_{n\in\N^+}\suf \cali{F}_{0}$ then $\curl{A_n^c }_{n\in\N^+}\suf \cali{F}_{0}$. Observe

\begin{align*}\cups{}{n\in\N^+}A_{n}=\dcups{}{n\in\N^+}A_{n}\del \cups{}{k\leq n-1}A_{k}=\dcups{}{n\in\N^+}A_{n}\cap \caps{}{k\leq n-1}A_{k}^{c}\end{align*}

Then $\caps{}{n\in\N^+}A_{n}^{c}\in\cali{F}$ by (2) and $\cups{}{n\in\N^+}A_{n}\in \cali{F}$ by (3) and previous observation. Thus $\cups{}{n\in\N^+}A_{n}\in\cali{F}_{0}$ as well. By (1), $\cali{F}_{0}$ is a $\s$-algebra containing all open subsets of $X$ so $\cali{F}\supf \cali{F}_{0}\supf \cali{B}_{X}$.$\square$

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The Eigenspace

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