A counterexample that shows addition and scalar multiplication is not enough for a vector space? According to Wikipedia, $(V, +, \cdot)$ is a vector space over field $\mathbb{F}$ if a list of axioms hold. But from my experience, what we usually do is to check whether closure under addition and scalar multiplication holds. I have two questions:


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*I'm looking for a counterexample for which closure under addition and scalar multiplication holds but it is not a vector space. Any idea?

*Why is it usually enough to check closure under addition and scalar multiplication?

 A: Consider the empty set, which is vacuously closed under any operation, but has no identity under addition.
A: Here's an example that has everything except for an identity element of scalar multiplication. Let $F$ be any field, and let $V = F^2$, with addition defined by member-wise addition. Define scalar multiplication by $ a(x, y) = (ax, 0) $.
In general, you do have to check all of the axioms, but when you have something that you think might be vector space, it usually has a particular structure. Almost always, $V$ is a subset of $F^I$, where $I$ is some index set, and vector addition and scalar multiplication are done element-wise.
Lemma: Suppose $F$ is a field, $I$ is a set, and $V \subseteq F^I$. Define:


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*$\mathbf{u} + \mathbf{v} = (u_i + v_i)_i$ for all $\mathbf{u}, \mathbf{v} \in V$

*$a \mathbf{u} = (au_i)_i$ for all $\mathbf{u} \in V$ and $a \in F$


If $V$ is closed under both addition and scalar multiplication, then it is a vector space.
Proof: Define the identity element of addition by $\mathbf{0} = (0)_i$. $0\mathbf{u} = \mathbf{0}$ for every $\mathbf{u} \in V$, so $\mathbf{0} \in V$ by closure. Define the inverse elements of addition by $-\mathbf{u} = -1 \cdot \mathbf{u}$, which exists in $V$ by closure. All other properties follow from the definitions and some algebra.
A: You can't get "Identity element of scalar multiplication" (i.e. $1 \cdot \vec v = \vec v$) from closure under addition and scalar multiplication.
