# What does proving a subset is a subspace accomplish?

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 $V$?

• So why is the subset $S=\{(1,0)\}$ of $\Bbb{R}^2$ closed under addition and scalar multiplication? It isn't because $(1,0)+(1,0)=(2,0)\notin S$. – Jyrki Lahtonen Mar 14 '18 at 6:25
• @JyrkiLahtonen that's a nice example. So a subset can contain some elements of the set but is not necessarily a subspace, but if a subset is shown to be a subspace, it's like a smaller defined vector space within the larger vector space? – mathguy Mar 14 '18 at 6:33
• Correct. One of the more economical ways of proving that a set is a vector space is to show that it is a subspace of a known vector space. – Jyrki Lahtonen Mar 14 '18 at 6:35
• @JyrkiLahtonen I understand why a subset needs to have closure properties to be a subspace, and it seems obvious to me that one wouldn't have to validate all the axioms again to prove a subset is a subspace, but I don't have an intuitive understanding as to why that is. Is it because every element of the vector space will satisfy the general axiom proofs therefore the elements in the subset will also satisfy them? – mathguy Mar 14 '18 at 7:42

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 $V$. $\square$

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.