p.18
This is the definition of subspace in his book:
1.32 Definition subspace
A subset $U$ of $V$ is called a subspace of $V$ if $U$ is also a vector space(using the same addition and scalar multiplication as on $V$).
and this is a proposition about subspace:
1.34 Conditions for a subspace
A subset $U$ of $V$ is a subspace of $V$ if and only if $U$ satisfies the following three conditions:additive identity
$0 \in U$closed under addition
$u, w \in U$ implies $u + w \in U$;closed under scalar multiplication
$a \in \mathbb{F}$ and $u \in U$ implies $a u \in U$.
I think there is a logical gap in his proof of the above proposition:
Proof:
If $U$ is a subspace of $V$, then $U$ satisfies the three conditions above by the definition of vector space.
I think $0 = 0_V \in U$ is not so obvious. My proof is the following:
My proof:
If $U$ is a subspace of $V$, then by the definition of vector space, $U$ is not empty.
Let $u$ be an element of $U$.
Because $U$ is a vector space, so $U$ is closed under scalar multiplication:
$0_V = 0 u \in U$.
Do you think there is a logical gap in his proof?
Best Answer
Your proof is correct. You can also use that $U$ has an additive neutral element $0_U$ with $0_U + 0_U = 0_U$ (as it is a vector space). This shows that $0_U$ is also the neutral element of $V$, so $0_V \in U$.
Whether to call this a logical gap is a matter of opinion. If you read mathmatical texts, you will often find steps or arguments you don't immediately understand. Then it is up to you to find the necessary points or to decide that the argumentation is wrong. Sometimes, this is really hard.
The phenomenon that a neutral element is already unique if it exists is something which comes up very often, and the arguments are very similar. So it is good to have the argumentation in mind anyway.