Vector Spaces over the Field of Real or Complex Numbers

Before we define a topological vector space, we will briefly recall what a vector space is and some related concepts to vector spaces.

Definition: A Vector Space over $\mathbf{F}$ (where $\mathbf{F}$ is either $\mathbb{R}$ or $\mathbb{C}$) is a set $E$ of elements called Vectors, equipped with two operations of addition between vectors, and scalar multiplication of elements of $\mathbf{F}$ with vectors, which satisfy the following properties: (1) $x + y = y + x$ for all $x, y \in E$. (2) $x + (y + z) = (x + y) + z$ for all $x, y, z \in E$. (3) There exists a vector $o \in E$ such that $x + o = x$ for all $x \in E$. (4) For each $x \in E$ there exists a vector $-x \in E$ with $x + (-x) = o$. (5) $(\lambda \mu)x = \lambda (\mu x)$ for all $\lambda, \mu \in \mathbf{F}$ and for all $x \in E$. (6) $(\lambda + \mu)x = \lambda x + \mu x$ for all $\lambda, \mu \in \mathbf{F}$ and for all $x \in E$. (7) $\lambda(x + y) = \lambda x + \lambda y$ for all $\lambda \in \mathbf{F}$ and for all $x, y \in E$. (8) $1x = x$ for all $x \in E$.

We will often just write "vector space" in place of "vector space over $\mathbf{F}$" unless we want to specify the field explicitly.

Definition: Let $E$ be a vector space. A Vector Subspace is a subset $M \subseteq E$ with the property that: (1) $(x + y) \in M$ for all $x, y \in M$. (2) $\lambda x \in M$ for all $\lambda \in \mathbf{F}$ and for all $x \in M$.

A subset $M$ of a vector space $E$ is sometimes said to be "closed under vector addition" if it satisfies condition (1), and is said to be "closed un scalar multiplication" if it satisfies condition (2).

Proposition 1: Let $E$ be a vector space and let $o \in E$ denote its zero vector. (1) $\{ o \}$ is a vector subspace of $E$. (2) If $\{ M_i : i \in I \}$ is a collection of vector subspaces of $E$ then $\displaystyle{\bigcap_{i \in I} M_i}$ is a vector subspace of $E$.

• Proof of (1): Clearly $o + o = o \in \{ o \}$ and for all $\lambda \in \mathbf{F}$, $\lambda o = o \in \{ o \}$ so that $\{ o \}$ is a subspace of $E$. $\blacksquare$

• Proof of (2): If $\displaystyle{x, y \in \bigcap_{i \in I} M_i}$ then $x, y \in M_i$ for all $i \in I$. Since each $M_i$ is a subspace of $E$ we have that $(x + y) \in M_i$ for all $i \in I$, so that $\displaystyle{(x + y) \in \bigcap_{i \in I} M_i}$. Also, since each $M_i$ is a subspace of $E$ we have that $\lambda x \in M_i$ for all $i \in I$, so that $\displaystyle{\lambda x \in \bigcap_{i \in I} M_i}$. So $\displaystyle{\bigcap_{i \in I} M_i}$ is a subspace of $E$. $\blacksquare$