I am having trouble figuring out how this notation works, specifically how the intersection relates to the rest of the summation. It's just stuck there after it.
I would greatly appreciate any help you can give in explaining it to me!
$$\left\vert \bigcup^{n}_{i=1} A_{i} \right\vert = \sum_{\substack{J \neq Ø \\ J \subseteq {[} n {]}}} (-1)^{\left\vert J \right\vert -1} {} \left\vert \bigcap_{i \in J} A_{i} {} \right\vert$$
On
The sum runs over every nonempty subset $J$ of $\{1, 2, 3, ..., n\}$. There will be exactly $2^n-1$ of these subsets, so there will be $2^n-1$ terms in the summation. I think this is most easily seen with an example. Say $n=2$. Then the sum would be:
$\left| \cup_{i=1}^n A_i \right| = (-1)^{|{\{1\}|-1}}|A_1| + (-1)^{|{\{2\}|-1}}|A_2| + (-1)^{|{\{1, 2\}|-1}}|A_1\cap A_2| = |A_1| + |A_2|-|A_1 \cap A_2|$
Hope this helps. If not, feel free to comment and I can try to clear things up.
$\textbf{Edit:}$
To clear up some questions of the OP, I'll include a simple example application of the inclusion-exclusion principle. I'll do an example in which $n=2$. That is, in which we want to determine the cardinality of the union of two sets.
Let $A_1 = \{1, 2, 3, 4, 5\}$, and $A_2 = \{3, 4, 5, 6, 7\}$. We wish to determine $|A_1 \cup A_2|$. We can easily see by inspection that $A_1 \cup A_2 = \{1, 2, 3, 4, 5, 6, 7\}$, so $|A_1 \cup A_2| = 7$. Therefore, we'll be able to check our answer after we use the inclusion-exclusion principle to compute $|A_1 \cup A_2|$.
By the inclusion-exclusion principle we have:
$$\left| \bigcup_{i=1}^2 A_i \right| = |A_1 \cup A_2| = \sum_{\substack{J \subseteq \{1, 2\} \\ J \ne \emptyset}} (-1)^{|J|-1}\left| \bigcap_{i \in J} A_i \right|$$
This means that for each nonempty subset $J$ of $\{1, 2\}$, we add a new term to our subset. The nonempty subsets of $\{1, 2\}$ are $\{1\}, \{2\},$ and $\{1, 2\}$. Therefore, the inclusion-exclusion principle gives:
$$\left| A_1 \cup A_2 \right| = (-1)^{|{\{1\}|-1}}\left|\bigcap_{i \in \{1\}} A_i\right| + (-1)^{|{\{2\}|-1}}\left|\bigcap_{i \in \{2\}} A_i\right| + (-1)^{|{\{1, 2\}|-1}}\left|\bigcap_{i \in \{1, 2\}} A_i\right| = (-1)^0 |A_1| + (-1)^0|A_2|+(-1)^1 |A_1 \cap A_2| = |A_1| + |A_2|+|A_1 \cap A_2|$$
Now, $A_1 = \{1, 2, 3, 4, 5\}$, $A_2 = \{3, 4, 5, 6, 7\}$, and $A_1 \cap A_2 = \{3, 4, 5, 6, 7\}$, so $|A_1| = |A_2| = 5$, and $|A_1 \cup A_2| = 3$. Plugging these into our most recent formula gives:
$$\left| A_1 \cup A_2 \right| = |A_1| + |A_2|+|A_1 \cap A_2| = 5 + 5 - 3 = 7$$
As stated above, we know that $A_1 \cup A_2 = \{1, 2, 3, 4, 5, 6, 7\}$, so $|A_1 \cup A_2| = 7$. Therefore, the inclusion-exclusion principle gave the correct answer! If you have any questions about this example, feel free to ask in the comments below. :)
On
Literally, the sum reads: Take the sum over the expression: $$(-1)^{|J|-1}\left|\bigcap_{i\in J}A_i\right|$$ for all non-empty subsets $J$ of the integers $\{1,2,\ldots,n\}$. The expression is the size of the intersection of the subsets $A_i$ where $i\in J$, negated when $|J|$ is odd. More intutively, you can read this as:
The size of the union of the $A_i$ is equal to the size of the intersection of an odd number of terms of $A_i$ (e.g. the size of each $A_i$ plus the size of any intersection of three $A_i$, ...) minus the size of the intersection of even numbers of terms from $A_i$ (e.g. the intersection of any pair). You could, equivalently read it as:
Sum up the sizes of each $A_i$. Now, subtract the size of the intersection of each pair. Now add the size of the intersection of each triplet, etc.
So, for $n=2$, the sum expands as: $$|A_i|+|A_j|-|A_i\cap A_j|$$
From the context, it seems $[n]$ is the power set of $\{1,2,3,....,n\}$, so its summing over all nonempty subsets $J$ of the power set. So, for subsets of length 1 (that's the $|J|$), you ADD the order of each $A_i$, then for subsets of length 2, it's saying to subtract each pairwise intersection, then add each triple-wise intersection, etc.