# If SS is an infinite σ\sigma algebra on XX then SS is not countable

I am going over a tutorial in my real analysis course. There is
an proof in which I don’t understand some parts of it.

The proof relates to the following proposition:

($$SS$$ – infinite $$σ\sigma$$-algebra on $$XX$$) $$⟹\implies$$ $$SS$$ is uncountable.

Proof:

Assume: $$S={Ai}+∞i=1S=\{A_{i}\}_{i=1}^{+\infty}$$. $$∀x∈X:Bx:=∩x∈AiAi\forall x\in X: B_{x}:=\cap_{x\in A_{i}}A_{i}$$. [Note: $$Bx∈SB_{x}\in S$$ $$⟸\impliedby$$ ($$BxB_{x}$$ – countable intersection].

Lemma: $$Bx∩By≠∅⟹Bx=ByB_{x}\cap B_{y}\neq\emptyset \implies B_{x}=B_{y}$$.

Proof(of lemma):

$$z∈Bx∩By⟹Bz⊆Bx∩Byz\in B_{x}\cap B_{y} \implies B_{z}\subseteq B_{x}\cap B_{y}$$.

1.$$x∉Bz⟹x∈Bx∖Bz∧Bx∖Bz⊂S∧Bx∖Bz⊂Bxx\not\in B_{z} \implies x\in B_{x}\setminus B_{z} \wedge B_{x}\setminus B_{z} \subset S \wedge B_{x}\setminus B_{z} \subset B_{x}$$
(contradiction:$$\space$$ definition of $$BxB_{x}$$) $$⟹\implies$$ $$Bz=BxB_{z}=B_{x}$$

2.$$y∉Bz⟹y∈By∖Bz ∧ By∖Bz⊂S ∧ By∖Bz⊂Byy\not\in B_{z} \implies y\in B_{y} \setminus B_{z} \space \wedge \space B_{y}\setminus B_{z} \subset S \space\wedge\space B_{y} \setminus B_{z}\subset B_{y}$$(contradiction: definition of $$ByB_{y}$$) $$⟹\implies$$ $$Bz=ByB_{z}=B_{y}$$ $$⟹Bx=By ◻\implies B_{x}=B_{y} \space \square$$

Consider: $${Bx}x∈X\{B_{x}\}_{x\in X}$$. If: there are finite sets of the
form $$BxB_{x}$$ then: $$SS$$ is a union of a finite number of disjoint sets
$$⟹\implies$$ $$SS$$ is finite $$⟹\implies$$ there is an infinite number of sets of
the form $$BxB_{x}$$. $$⟹\implies$$ $$|⋃i∈A⊆NBxi|≥ℵ0|\bigcup\limits_{i\in A \subseteq\mathbb{N}}B_{x_{i}}| \geq \aleph_{0}$$.(contradiction) $$◻\square$$

There are couple of things I don’t understand in this proof:

1. Why the fact that we found a set ($$Bx∖BzB_{x}\setminus B_{z}$$) in $$SS$$
containing $$xx$$ and is strictly contained in $$BxB_{x}$$ a contradiction
?

2. Why if there are only a finite number of different sets of the
form $$BxB_{x}$$ then $$SS$$ is a union of a finite number of disjoint
sets and is finite ?

1. Because $B_x$ is supposed to be the intersection of all measurable sets containing $x$, but you’ve found a measurable set containing $x$ strictly inside $B_x$.
2. Because for any measurable set $T$, we have $T=\bigcup_{x\in T}B_x$. Thus, if there are $n$ distinct sets of the form $B_x$, then there are at most $2^n$ elements of $S$.