Topological linear space

Question

Let $E$ be a topological linear space. A subset $M$ of $E$ is said to be bounded if $\forall U\in I(0)$ there exist a scalar $\lambda$ such that $M\subseteq \lambda U$, where $I(0)$ denotes the family of neighbourhoods of $0$. How to prove that each point of $E$ is bounded?

Answer 1

Suppose $M = \{x\}$ for some $x \in E$.

Let $m: \Bbb F \times E \to E$ be the continuous scalar multiplication map, and it's well known that $m(0,x)=0$. So if $U \in I(0)$, we can find $V$ open in $\Bbb F$ such that $cx \subseteq U$ for all $c \in V$, by continuity.

Pick any $c \neq 0$ in this $V$ (I need that $\Bbb F$ is not discrete so that this is possible) and define $\lambda=\frac{1}{c}$. Then $M \subseteq \lambda U$ as $cx \in U$ implies $x = \lambda(cx) \in \lambda U$.

Answer 2

Let $x\in E$, set $M=\{x\}$. Let $(\lambda_n)\subset\mathbb{F}$ be any sequence of scalars such that $\lambda_n\to0$. Since $E$ is a TVS, we have that $\lambda_nx\to0$ in $E$, so for any $U\in I(0)$ there exists $n_{0,U}\in\mathbb{N}$ so that $\lambda_nx\in U$ for all $n\geq n_{0,U}$. In particular, $x\in \frac{1}{\lambda}U$, where $\lambda=\lambda_{n_{0,U}}$, i.e. $M\subset \frac{1}{\lambda}U$.