Properties about topological vector spaces Let $E$ be a topological vector space. First I want to prove that, given a $V \subset E$ balanced and $\lambda>0$ then
$$
\lambda V \subset \beta V, \: \forall \;\lambda< \beta. \tag{1}.
$$
For this I tried the following: let $\lambda,\beta>0$ such that $\lambda<\beta$ and let $x \in  \lambda V$. Thus, there exists $v \in V$ satisfying $x=\lambda v$. Since $V$ is balanced, then
$$\frac{\lambda}{|\lambda|}v = \frac{\lambda}{\lambda}v=v \in V \Rightarrow \frac{1}{\lambda}x \in V \Rightarrow \frac{\beta}{\lambda}x \in \beta V.$$
But, how can I conclude that $ x \in \beta V $?
I also want to prove that: given $U\subset E$ a neighborhood of $0 \in E$ such that $U$ is open and balanced, then
$$H:= \bigcup_{n=1}^{\infty}nU=E.\tag{2}$$
It's clear that $H \subset E$. But the converse inclusion , that is, given $x \in E$ there exists $n_0 \in \mathbb{N}$ so that $x=n_0U$, I couldn't prove.
Are my ideas right? Any suggestion?
Remember that: a subset $A$ of a vector space $X$ is said to be balanced if for every $x \in A$ and every $\lambda \in \mathbb{C}$, $|\lambda|\leq 1$, we have $\lambda x\in A$.
 A: For the first bit: If $x=\lambda v\in \lambda V$, then $x=\beta \frac{\lambda}{\beta}v \in \beta V$, as $\vert \lambda/\beta\vert \le 1$ and $V$is balanced.
For the second question,balanced is not needed. If $x\in E$, the map $\alpha \mapsto \alpha x$ (defined on  the base field) is continuous and thus, if $n\rightarrow \infty$, we have $x/n\rightarrow 0$. In particular, the sequence eventually lies in $U$ (the open neighbourhood of $0$) and for some $n$ we have $x/n\in U$.
A: The implication
$$0 < \lambda < \beta \implies \lambda V \subseteq \beta V$$
is not very hard:
Since $V$ is balanced and $\lambda/\beta < 1$, we have $\frac{\lambda}{\beta} V \subseteq V$. Multiplying both sides with $\beta$ gives us what we want.
For your second question, let us show that $E= \bigcup_{n=1}^\infty nU=:H$.  By continuity of scalar multiplication, we have the following property:
$$\forall e \in E: \exists \delta  > 0: \forall 0 < \lambda \leq \delta: \lambda e \in U$$
Given $e \in E$, choose a $\delta$ as above. Then choosing $n$ zo large that $1/n \leq \delta$, we obtain
$$\frac{1}{n}e \in U \implies e \in nU$$
A: partial answer
$V$ being balanced means that $\forall |\mu|< 1$, $\mu V\subset V$. Now use that $\frac{\lambda}{\beta}< 1$
