Topological Vector Space as Union of Dilations of Open Neighborhood of Origin

Theorem

Let $\Bbb F \in \set {\R, \C}$.

Let $X$ be a topological vector space over $\Bbb F$.

Let $V$ be an open neighborhood of ${\mathbf 0}_X$.

Let $\sequence {r_n}_{n \mathop \in \N}$ be a sequence of positive real numbers such that $r_n \to \infty$.

Then:

$\ds X = \bigcup_{n \mathop = 1}^\infty r_n V$

where $r_n V$ denotes the dilation of $V$ by $r_n$.

Clearly:

$\ds \bigcup_{n \mathop = 1}^\infty r_n V \subseteq X$

We show that:

$\ds X \subseteq \bigcup_{n \mathop = 1}^\infty r_n V$

Let $x \in X$.

From the definition of a topological vector space, the mapping $K \times X \to X$ defined by $\tuple {\lambda, x} \mapsto \lambda x$ is continuous.

Hence, the set:

$U = \set {\lambda \in \Bbb F : \lambda x \in V}$

Since $0 x = {\mathbf 0}_X$ and $V$ is an open neighborhood of ${\mathbf 0}_X$, we have:

$U$ is an open neighborhood of $0 \in \Bbb F$.

So, there exists $\epsilon > 0$ such that if $\cmod \lambda < \epsilon$, we have $\lambda x \in U$.

From the definition of a real sequence tending to $+\infty$, there exists $N \in \N$ such that:

$\ds r_n > \frac 1 \epsilon$ for $n \ge N$.

Then:

$\ds 0 < \frac 1 {r_N} < \epsilon$

so that:

$\dfrac 1 {r_N} x \in V$

giving:

$x \in r_N V$

So, we have:

$\ds x \in \bigcup_{n \mathop = 1}^\infty r_n V$

Since $x \in X$ was arbitrary we have:

$\ds X \subseteq \bigcup_{n \mathop = 1}^\infty r_n V$

giving:

$\ds X = \bigcup_{n \mathop = 1}^\infty r_n V$

$\blacksquare$