Suppose U1,…,UmU_1,\dots,U_m are finite dimensional subspaces of VV. Prove that U1+⋯+UmU_1 + \dots + U_m is finite dimensional and

dim⁡(U1+⋯+Um)≤dim⁡U1+⋯+dim⁡Um\dim(U_1+\dots+U_m) \le \dim U_1 + \dots + \dim U_m

Let u1j,…,udim⁡Ujju^j_1,\dots,u^j_{\dim U_j} be a basis of UjU_j for all 1≤j≤m1 \le j \le m. Concatinating all the bases gives a list of length dim⁡U1+⋯+dim⁡Um\dim U_1 + \dots + \dim U_m which spans U1+⋯+UmU_1 + \dots + U_m thus it's finite dimensional. Then apply 2.31 to remove dependent vectors until we have a basis for U1+⋯+UmU_1 + \dots + U_m. Thus

dim⁡(U1+⋯+Um)≤dim⁡U1+⋯+dim⁡Um\dim(U_1+\dots+U_m) \le \dim U_1 + \dots + \dim U_m