This is a statement which I found without proof and maybe it's obvious, but I can't understand why it should be true:
Let $X$ be a locally convex vector space, $\mathcal{W}$ a neighbourhood basis of zero, consisting of balanced, convex, closed sets. Then $\bigcup_{W \in \mathcal{W}} W^\circ = X' $.
$W^\circ$ does mean the polar of $W$ and $X'$ the topological dual space, so all linear, continuous functionals on $X$. So actually my problem is: Why is a function $f$ from the algebraic dual space $X^*$ continuous if and only if there is a $W \in \mathcal{W}$ such that $|<x,x^*>| \le 1$ with $x \in W$...?
The set $U=\{a\in\mathbb K:|a|\le 1\}$ (either an interval or a disc depending on the field $\mathbb K$) is certainly a neighbourhood of $0$ in $\mathbb K$. Since every continuous linear map $f$ maps $0\in X$ to $0\in\mathbb K$, there is a $0$-neighbourhood $W$ in $X$ with $f(W)\subseteq U$, hence $f\in W^\circ$.
You might want to try the other implication yourself.