Why we take V a non empty set in vector space?

Question

In functional analysis why we take V as a non-empty set? And when we say that a vector space V over F.can we say that F is any field?

Answer 1

If $V$ is a vector space, then $V$ is never empty, since $V$ contains at least one element: the "null-vector".

In functional analysis the vector spaces are usually normed spaces or topological vector spaces over the field $F \in \{ \mathbb R, \mathbb C\}$.

Answer 2

$F$ can be any field, yes. However, it is usually the case that $F $ is either the field of complex numbers or the field real numbers. As the answer of @Fred pointed out, by definition: a vector space $V $ has to contain the 'null vector'. Therefore, its nonempty by definition.

Answer 3

In functional analysis why we take V as a non-empty set?

Really there's no fundamental reason. It's just that the special case where $V$ is empty would not add anything (it obviously wouldn't be a very interesting case) and it would add special cases to many theorems to deal with this pesky "vector space". It would be not unlike taking 1 as a prime, or deciding that 0 is neither even nor odd. Fundamentally unimportant but it makes many theorems uglier.

Answer 4

If $V$ were the empty set, it would not be a vector space, because a vector space is an abelian group, and an abelian group has a neutral element, and the empty set clearly contains no neutral element.

For a vector space, $F$ can be any field. However you asked about functional analysis, not linear algebra, and in functional analysis we want to do limits, therefore we want the field to be topological (so we can define limits). We'd also like to have as many sequences as possible to have limits; the best way to ensure that is if the field is also a complete metric space. This pretty much restricts the field to be either $\mathbb R$ or $\mathbb C$.