every compact subset of a TVS is bounded I'm self-studying Functional Analysis. The following is Exercise 4.2.4 of Conway's Functional Analysis.
Let $X$ be a topological vector space. Show that every compact subset of $X$ is bounded.
For this I just know every open cover of compact set $K$ has a open finite sub cover; but I can not find a relation to show that it's bounded.
Please help me. Thanks in advance.  
 A: Suppose that $C\subseteq \mathscr X$ is compact, $U\subseteq\mathscr X$ is open, and $0\in U$. By the continuity of scalar multiplication at the point $(0,0)\in\mathbb F\times\mathscr X$, there exists an open set $W\subseteq\mathscr X$ such that $0\in W$ and some $t>0$ such that $s W\subseteq U$ for any $s\in(0,t)$. Let $$V\equiv\bigcup_{s\in(0,t)}s W\subseteq U.$$
Being the union of open sets, $V$ is an open set that contains $0$ and $\alpha V\subseteq V$ for any $\alpha\in(0,1)$. 
Fix $x\in C$. The map $z\mapsto zx$ is continuous from $\mathbb F$ to $\mathscr X$ (as the composition of $z\mapsto (z,x)$ from $\mathbb F$ to $\mathbb F\times\mathscr X$—which is the product of the identity map and a constant map—and $(z,x)\mapsto zx$ from $\mathbb F\times\mathscr X$ to $\mathscr X$), so there is an open set $T\subseteq\mathbb F$ containing $0$ such that if $z\in T$, then $zx\in V$.
Pick any $t_x\in T$ such that $t_x\in(0,t)$, so that $x\in (1/t_x)V$. The set $(1/t_x)V$ is open by virtue of continuity of scalar multiplication. (Indeed, $(1/t_x)V$ is the preimage of $V$ under the continuous map $y\mapsto t_x y$ from $\mathscr X$ to $\mathscr X$.) Therefore,
$$C\subseteq \bigcup_{x\in C}\left(\frac{1}{t_x}\right)V.$$
By compactness,
$$C\subseteq\bigcup_{n=1}^N\left(\frac{1}{t_{x_n}}\right)V,$$
for some $n\in\mathbb N$ and $\{x_n\}_{n=1}^N\subseteq C$.
Let $t^*\equiv\min\{t_{x_1},\ldots, t_{x_N}\}/2$. Then,
$$C\subseteq\frac{1}{t^*}\bigcup_{n=1}^N\left(\frac{t^*}{t_{x_n}}\right)V\subseteq\frac{1}{t^*} V\subseteq \frac{1}{t^*} U.$$
It follows that $t^* C\subseteq U$, showing that $C$ is bounded.
A: Hint: If U is a balanced neighborhood of $0$, then $\bigcup_n nU=X$.
