Is the $\pi$-system generated by a topology the topology itself?

Question

I was studying the introduction for my probability course, and came to the definition of a $\pi$-system.

Let $C \subset \mathcal{\Omega}$. $C$ is a $\pi$-system if $\forall A, B \in C \implies A \cap B \in C$.

But from the definition of a topology, does that mean that if $\mathcal{T}$ is a topology, then $\pi(\mathcal{T}) = \mathcal{T}$?

Where $\pi(\mathcal{T})$ is the $\pi$-system generated by $\mathcal{T}$.

If this is the case, I would ask a less rigorous question: Is the $\pi$-system the probability theory equivalent of a topology?

EDIT: The definition I am using for the $\pi$-system generated by $\mathcal{T}$ is the smallest $\pi$-system that contains the topology $\mathcal{T}$

Answer 1

Denote the topology in the question with $\mathcal{T}$. We want to prove that

$$\pi(\mathcal{T}) = \mathcal{T}$$

Clearly, we have $\mathcal{T} \subseteq \pi( \mathcal{T}).$

The other inclusion $\pi (\mathcal{T}) \subseteq \mathcal{T}$ is immediate, since $\mathcal{T}$ is a $\pi$-system that contains $\mathcal{T}$, and we then use minimality.