Proof that sum of two subspaces is another subspace

Question

$U_1,U_2$$⊂V$ be subspaces of V (a vector space). Define the subspace sum of $U_1,$ and $U_2$ be defined as the set:

$U_1 + U_2$ $=$ {$u_1 + u_2 : u_1 ∈ U_1, u_2 ∈ U_2$}.

Let $A$ denote the set $U_1+ U_2$

A is a subspace if a meets all the criteria of a subspace, that is, $0∈A$, it remains closed under addition, and it remains closed under multiplication.

Since $U_1$ is a subspace, by definition of subspaces it contains $au_1$ ($a∈R$), $0$ when a equals zero, and $u_1 + w_1$ (when $w_1∈U_1$).

Since $U_2$ is a subspace, by definition of subspaces it contains $au_2$ ($a∈R$), $0$ when a equals zero, and $u_2 + w_2$ (when $w_2∈U_2$).

$0u_1 + 0u_2 = 0(u_1 + u_2) = 0$; is an element of $A$

$au_1 + au_2 = a(u_1 + u_2)$; is an element of $A$

$(u_1 + u_2) + (w_1 + w_2) = (u_1 + w_1) + (u_2 + w_2)$ is an element of $A$

Q.E.D.

This is my proof, is it correct logically, symbolically etc., does it fall short of clarity, format/structure etc.?

In general what tips would you give to a young (a.k.a. not very mathematically mature) self-learner to improve their proofs. More specifically, what should I work on based on my proof.

Answer 1

Yes your proof is fine as a minor issue I would prefer to present the second and third properties in that way

• $u_1 + u_2\in U_1 + U_2 \implies a(u_1 + u_2)=au_1 + au_2$ with $au_1\in U_1 + U_2$ and $au_2\in U_2 $

and

• $(u_1 + u_2) + (w_1 + w_2)\in U_1 + U_2 \implies (u_1 + u_2) + (w_1 + w_2)=(u_1+w_1)+(u_2+w_2)$ and $(u_1+w_1)\in U_1$, $(u_2+w_2)\in U_2$