What does proving a subset is a subspace accomplish?

Question

If $V$ is a vector space that has closure properties and satisfies the axioms and $S$ is a subset of $V$, why wouldn't $S$ always have closure under addition and scalar multiplication (which are required to show $S$ is a subspace) because since $S$ is a subset of $V$, doesn't that mean $S$ would have the same properties as $V$?

Answer 1

Take the trivial example of $V=\mathbb R$ as vector space over $\mathbb R$. Take $S$ to be any set other than $\{0\}$ and $\mathbb R$ itself. Then $S$ is not a subspace of $V$.

Proof:

If $S$ does not contain $0$, it does not contain a neutral element of addition (this especially also applies to the empty set). Thus it cannot be a vector space, and in particular no subspace of $V$.

If $S$ contains any $x\in\mathbb R\setminus\{0\}$, but there exists an $y\in\mathbb R\setminus S$, then be $\lambda=y/x\in\mathbb R$. Then $\lambda x = y\notin S$ despite $x\in S$ and $\lambda\in\mathbb R$, therefore scalar multiplication is not closed in $S$. Thus $S$ again is no subspace of $V$. $\square$

Answer 2

No, because it could be that $v, w \in S $ but $v+w \not\in S $ or $\lambda v \not\in S$.

Note that you have to prove both closure properties: one does not entail the other.

Here are some examples:

Take $V = \mathbb{R}^2 $ . Now, $ \{(x,y)\in \mathbb{R}^2 \vert x\in \mathbb{Z}, y\in \mathbb{Z} \} $ . This subset is closed under addition but not under scalar multiplication (for a number in $\mathbb{R}$, which is our field).

Then consider $\{(x,y) \vert x=0 \vee y=0 \} $. This set is closed under scalar multiplication but not under addition.

Lastly, consider $\{(x,y) \vert x^2+y^2=1 \} $. This set is not closed under any of the two operations.