The $\sigma$-Algebra generated by disjoint sets has the same cardinality as the power set of those disjoint sets Is it true that the $\sigma$-Algebra generated by disjoint sets has the same cardinality as the power set? Intuitively it is, but how can it be proven?
 A: It is not true. First off I think you have forgotten the assumption that your disjoint sets cover the original space, if this is not the case consider for example $X=\{1,2,3\}$ and take the $\sigma$-Algebra generated by $\{1\}$ and $\{2\}$. Since $X-(\{1\}\cup\{2\})=\{3\}$ lies in it you have that all subsets of $X$ lie in it, which has cardinality $8$ whereas the power set of $\{\{1\},\{2\}\}$ has cardinality $4$.
In the case that your generating set $\mathcal A=\{U_i\mid i\in I\}$ covers $X$, we will explicitly construct the $\sigma$-Algebra. In this case there exists a bijection to $\sigma(\mathcal A)$ from
$$\mathcal{C}_I:=\{J \subset I \mid J \text{ is countable}\}\cup \{K \subset I \mid I-K \text{ is countable}\}$$
Call here the first set $\mathcal J$ and the second set $\mathcal K$. We give the map from $\mathcal C_I$ to $\sigma(\mathcal A)$ by sending elements $J$ of $\mathcal J$ to $\bigcup_{j\in J}U_j$ and sending elements $K$ of $\mathcal K$ to $\bigcup_{k\in K }U_k=X-\bigcup_{i\in I-K}U_i$.
In the case that $I$ is countable $\mathcal J$ and $\mathcal K$ are the same sets, but they are also sent to the same sets so the map is well defined in this case. In the case that $I$ is uncountable the sets $\mathcal J$ and $\mathcal K$ are disjoint and the map is also well defined.
The image of the map is clearly a subset of $\sigma(\mathcal A)$ and the map is also clearly injective, since the $U_i$ are all pairwise disjoint. If we now show that the image is a $\sigma$-Algebra we have that it is a bijection.
To this end note that the set is closed under complements and that:
$$\bigcup_n\bigcup_{i\in I_n} U_i=\bigcup_{i\in \bigcup_nI_n}U_i$$
We take $I_n\in\mathcal C_I$. If one of the sets $I_n\in\mathcal K$ it has countable complement then clearly $\bigcup_nI_n$ has countable complement. If no set $I_n\in\mathcal K$ then $\bigcup_nI_n$ is a countable union of countable sets and is countable. So the image is closed under countable unions.
Now:
$$\bigcap_n\bigcup_{i\in I_n}U_i=\bigcup_{i\in \bigcap_nI_n}U_i$$
If one $I_n\in\mathcal J$ that then $\bigcap_n I_n$ must be countable. If no $I_n\in\mathcal J$ then this is a countable intersection of sets with countable complement, so it also has a countable complement. So the image is also closed under countable intersections.
Thus we have seen that case $\sigma(\mathcal A)$ can be identified with the "countable powerset" of $\mathcal A$, ie the set of countable subsets and subsets with countable complement. This does not need to have the same cardinality as the powerset. Though if $\mathcal A$ is countable then it does!
