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Question Does there exists an infinite $\sigma$-algebra which has only countably many members?

Solution Let $\mathcal{A} = \{A_1, A_2, \dots \}$ be the countable $\sigma$-algebra of $X$.

Define the equivalence relation $x \sim y \iff (x \in A_i \leftrightarrow y \in A_i)$. Then $X / \sim$ is a set of distinguishable elements of $X$. $X / \sim$ is at least countable, because otherwise $\mathcal{A}$ will be finite.

For every distinct $x, y \in X/\sim$, there exists $A \in \mathcal{A}$ such that $x \subset A, y \not\subset A$. Then because $\mathcal{A}$ is countable, we can construct every $x \in X/\sim$ with countable number of substitution.

Let $Y$ be the countable subset of $X/\sim$. $Y \subset \mathcal{A}$ and $Y$ is countable, so the uncountable set $\mathcal{P}(Y)$ lies inside $\mathcal{A}$ which leads to contradiction.

How does it look? If it is wrong, can it be fixed with minor changes? Or is it completely wrong?

Thanks in advance.

SU Lee
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1 Answers1

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I only answer the question given in your post. We're going to show that sigma algebras are either finite or not countable.

Lemma: Let $\mathcal{A}$ be an infinite $\sigma$-algebra on some set $X$. Then there exists an infinite amount of disjoint sets in $\mathcal{A}$.

Proof: We'll construct our sets by hand. Let $(A_i)_{i\in\mathbb{N}}\in\mathcal{A}^\mathbb{N}$ and for all $(\omega_i)\in\{0,1\}^\mathbb{N}$ such that:

$$B_{\omega}=\bigcap_{i\in\mathbb{N}}B_{\omega_i,i}$$

Where $B_{1,i}=A_i$ and $B_{0,i}=X\backslash A_i$. Then $(B_\omega)_{\omega\in\{0,1\}^\mathbb{N}}$ are all disjoint one to one and are all in $\mathcal{A}$ as it is a $\sigma$-algebra. Unfortunately, many of them are empty, and so we'll have to show that an infinite amount of them aren't. To do so, notice how:

$$\forall i\in\mathbb{N},\ A_i=\bigsqcup_{\omega,\ \omega_i=1}B_\omega$$

And so if only a finite amount of $(B_\omega)$ were non empty we'd only be able to reconstruct a finite amount of $(A_i)_{i\in\mathbb{N}}$, proving the lemma.

We can now show that all $\sigma$-algebras on some set $X$ are either finite our uncountable.

Proof: I'm only going to treat the infinite case. Thanks to our lemma, if we suppose $\mathcal{A}$ infinite then there exists a sequence of non empty one to one disjoint elements $(\Omega_i)_{i\in\mathbb{N}}$. Now consider $A\subset\mathbb{N}$ and:

$$D_A=\bigsqcup_{n\in A}C_n$$

Each $D_A$ is different (take the intersection of two distinct ones) and $\mathcal{P}(\mathbb{N})\cong\mathbb{R}$, proving the theorem; $\mathcal{A}$ isn't countable.

J.J.T
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