TITLE: Disjoint Union of Subsets and Direct Sum of Subspaces (Clarify Explanation)
QUESTION [7 upvotes]: Can someone clarify for me the side note found in Axler's Linear Algebra Done Right, that is: 
Direct sums of subspaces are analogous to disjoint union of subsets. 
I am not sure what exactly is a disjoint union of subsets (having checked online definitions) and thus its relation to direct sums. 
Thanks.

REPLY [3 votes]: For finite subsets of some set $U$, one has 
$$
  \#(A_1\cup A_2\cup\cdots\cup A_n)\leq\#A_1+\#A_2+\cdots+\#A_n,
$$
with equality if and only if the sets $A_1,A_2,\ldots,A_n$ are all disjoint,which is a condition one can test by considering the $A_i$ pairwise (i.e., one may test that $\#(A_i\cap A_j)=0$ whenever $i\neq j$). For finite dimensional subspaces of some vector space $V$ one has$$
  \dim(V_1+V_2+\cdots+V_n)\leq\dim V_1+\dim V_2+\cdots+\dim V_n,
$$
with equality if and only if the subspaces $V_1,V_2,\ldots,V_n$ form a direct sum $V_1\oplus V_2\oplus\cdots\oplus V_n$ (this is just notation to indicate the sum is direct); this is a condition one cannot test by considering the $V_i$ pairwise (i.e., it does not suffice to test that $\dim(V_i\cap V_j)=0$ whenever $i\neq j$).
So direct sums have something similar to disjoint unions, but not everything.