Why is the dual space with the weak*-topology a topological vector space? In my lecture-notes on functional analysis I've found the fact that the dual-space $X^*$ with the weak*-topology of a real vector-space $X$ is a topological vector-space. 
I've tried to prove it, but I couldn't find the right way to attack it. Does anyone know how to do it?
Thanks for any help!
 A: You need to prove that two maps
$$
m : X^{\ast} \times X^{\ast} \to X^{\ast} \text{ given by } (f,g) \mapsto f+g
$$
and
$$
s : k\times X^{\ast} \to X^{\ast} \text{ given by } (\alpha, f) \mapsto \alpha f
$$
are continuous. Let us prove it for $m$, and leave $s$ for you to tackle.
Suppose $W \subset X^{\ast}$ is weak-$\ast$ open, then we want to show that $m^{-1}(W)$ is open. Choose $(f,g) \in m^{-1}(W)$, then we want to show that $\exists U,V \subset X^{\ast}$ weak-$\ast$ open such that
$$
(f,g) \in U\times V \text{ and } U\times V \subset m^{-1}(W)
$$
Since $h:= f+g \in W$ and $W$ is open, $\exists \epsilon >0, x_1,x_2,\ldots, x_n \in X$ such that
$$
\{k \in X^{\ast} : |k(x_i) - h(x_i)| < \epsilon \quad\forall 1\leq i\leq n\}
$$
So let
$$
U = \{f_1 \in X^{\ast} : |f_1(x_i) - f(x_i)| < \epsilon/2 \quad\forall 1\leq i\leq n\}
$$
and
$$
V = \{g_1 \in X^{\ast} : |g_1(x_i) - g(x_i)| < \epsilon/2 \quad\forall 1\leq i\leq n\}
$$
Then for any $(f_1,g_1) \in U\times V$, we have
$$
|(f_1+g_1)(x_i) - h(x_i)| < \epsilon \quad\forall 1\leq i\leq n
$$
Hence
$$
U\times V \subset m^{-1}(W)
$$
and so $m$ is continuous.
