I had some doubt with my proof, but I'll list the question here along with the proof:
Claim: Show that the cardinality of a finite $\sigma$-algebra $\mathfrak{M}$ on a set $X$ is $2^n$ for $n \in \mathbb{N}$. Describe the exponent $n$ in terms of $\sigma$-algebra.
Proof: For each member in $\mathfrak{M}$, we can pair it with its complement by properties of $\mathfrak{M}$. This pairing gives a specific partition. Using this idea, we can pick any combination of members of $X$ such that these sets of members of $X$ forms a partition of $X$. Just observing the collection of partitions of $X$, we consider the partition which has the most cells by superimposing all of the partition on top of each other. So suppose there are $n$ cells for a specific partition of $X$. Let $S$ be the set that contains these $n$ cells. So $\mathcal{P}(S) = 2^n$, where $n$ is the maximum number of cells one can achieve through all possible partition of $X$. From $\mathcal{P}(S)$, we get the other partition of $X$. Hence, $S$ generates $\mathfrak{M}$.
On a similar note, there's a question that is similar to the one I am posing:
If we are given any infinite ($|\mathfrak{M}| =$ is infinite) $\sigma$-algebra $\mathfrak{M}$ on set $X$, then there is a subset with cardinality of the real numbers $2^{\aleph_{0}}$.
Proof: I don't think my argument would work in the infinite case, but I'll give it a go. So I thought that $\mathfrak{M}$ has infinitely many partition of $X$, so if we were to countably infinitely take intersection of the partitions of $X$, we get a countably infinite cells of a partition of $X$. Using the argument by the previous problem, we take the power set of the natural number, which gives us $2^{\aleph_{0}}$. For some reason, I am unsure how this infinite case would work out.
-Thanks in advance