is the (countable) partition generated $\sigma$-algebra a complete lattice?

Question

Let $T$ be a (countable) partition of $X$ and let $\sigma(T)$ be the generated $\sigma$-algebra of our interest.

I'm trying to figure out whether $\sigma(T)$ is a complete lattice?

Def. a complete lattice is a partially ordered set in which all subsets have both a supremum (join) and an infimum (meet). E.g. Any power set is a complete lattice.

I was thinking something along the line of Since there is a one-to-one correspondence between a powerset and $\sigma(T)$, and the powerset is a complete lattice, then $\sigma(T)$ is a complete lattice.

Can you please provide me with some help on the subject?

Answer 1

Since $T$ is a countable partition $\sigma(T)$ will have as elements exactly the sets that can be written as a union of elements of $T$.

You can prescribe $F:\wp(T)\to\sigma(T)$ by $S\mapsto\cup S$.

(Here $\cup S:=\bigcup_{R\in S}R$)

It is not really difficult to verify that $F$ is bijective and order preserving, and also that $F^{-1}$ is order preserving.

That makes $F$ an isomorphism in the category of partial ordered sets.

So proving that $\langle\sigma(T),\subseteq\rangle$ is a complete lattice is actually the same as proving that $\langle\wp(T),\subseteq\rangle$ is a complete lattice.

Answer 2

Since the partition is countable, let's say $X = \bigcup_{n \in \mathbb{N}} S_n$ you can define $$ \varphi : \sigma(T) \to 2^\mathbb{N} \quad \varphi(S_n) = \{n\} \quad \varphi \left(\bigcup_{i \in \mathbb{N}} X_i \right) = \bigcup_{i \in \mathbb{N}} \varphi(X_i) $$

It is easy to see that this function is well defined bijection, since each set in $\sigma(T)$ is a countable union of the sets in your partition, and it preserves the partial relation. Thus $\sigma(t)$ and $\mathbb{N}$ have the same lattice structure, and so $\sigma(T)$ is complete.

remark this is just a sketch, if you need more detail, just ask