$\sigma$-algebra and disjointness Let $X$ be a space and $S$ be a $\sigma$-algebra on $S$. We often try to find disjoint sets that cover $X$ (a partition of $X$) that "generates" $S$, but what is the importance of disjointness? Why is it important for the sets not to intersect? Can we not generate a $\sigma$-algebra using sets that intersect? 
 A: You perfectly well can. For example, it is often nice to note that $[x,\infty)$ generates the Borel $\sigma$ algebra on $\mathbb{R}$. Disjointness is nice once you start introducing measures, as $\mu(E \cup F) = \mu(E) + \mu(F)$ (where $\mu$ is a measure on a $\sigma$-algebra $\mathcal{M}$ and $E, F \in \mathcal{M}$) provided $E \cap F = \emptyset$ (and it is not true in general if they do intersect). Hence having a disjoint generating set can be helpful when you are trying to define measures and things like that.
A: Here is an answer based on the explanation you gave as a comment to Keefer Rowan's answer.
Why is it better to have disjoint sets $A_1, A_2, \ldots$ to show your sigma algebra is uncountable:
If they are disjoint you can easily see that your sigma algebra has at least the same cardinality as the power set of $\Bbb{N}$ denoted by $\mathcal{P}(\Bbb{N})$. 
Because for each $I,J \in \mathcal{P}(\Bbb{N})$ with $I \not= J$ it then follows that $$\bigcup_{i \in I} A_i \not= \bigcup_{j \in J} A_i$$
So for each element from  $\mathcal{P}(\Bbb{N})$ you will find a unique element in your sigma algebra.
This is not the case if your $A_i$ are not disjoint! So you cannot conclude anymore that your sigma algebra has at least the same cardinality as $\mathcal{P}(\Bbb{N})$ so you can get no information out if it!
To see this consider e.g. $A_1 = \{1,2\}, A_2 = \{3\}, A_3 = \{1,2,3\}$ then for $$I = \{1,2\}, J = \{3\}$$ it will hold $$\bigcup_{i \in I} A_i = \bigcup_{j \in J} A_i$$
So the definiton above defines a mapping from $\mathcal{P}(\Bbb{N})$ into your sigma algebra by $$I \mapsto \bigcup_{i \in I} A_i$$
If the $A_i$ are disjoint this mapping is injective, if they are not, it's not. And if you've found an injective mapping from one set to another one you can conclude, that the cardinality of the first one is less or equal to the second one.
