Suppose the following: $$N=[3,\infty)\cap \Bbb Z$$ $$n \in N$$ $$N'=[1,n] \cap \Bbb Z$$ $$i,j,k \in N'$$ $$|S| \in N \setminus N'$$ $S_{i} \cap S_{j}=\emptyset $ when $i \not =j$, and $$S:= \bigcup _{k \in N'} S_k$$ Then, is the following true? $$|S|=\sum_{k \in N'}|S_k|$$ I was thinking that it was because the intersection of any $S_i$ and $S_j$ is empty for $i \not =j$, but I am not sure how to prove it. (Sorry if my notation is confusing\wrong, I am fairly new to set theory.)
Edit: Translating it into words: If $S$ is a finite, non-empty set, partitioned into $n$ strict subsets, such that the intersection of any two of these subsets is $\emptyset$, is the cardinality of $S$ just the sum of the cardinalities of the subsets?
Yes, it is true. You should know, and it sounds like you do, that the union of two disjoint subsets has size the sum of the sizes of the subsets. For three subsets, just add two of them to make one set, then add in the third. Proceed by induction of the number of subsets. Many properties are like this, if you prove it for a pair you have proven it for all finite collections.