weak-* topology What is the definition of weak-*topology ($\sigma(X^*,X)$)? What is the intuition standing behind its definition (apart from easier extraction of finite subcovers - Banach - Alaoglu theorem)?
I start with weak topology. So let $X, Y_j$ be topological spaces and let $f_j\colon X\to Y_j$, $j\in J$. Consider a collection
$$\mathcal{O}=\{\bigcap_{j=1}^k f_{j}^{-1}(O_j)\mbox{-open in}\,X,\, k\in\mathbb{N},\,O_j\, \mbox{open in}\, Y_j\}.\qquad (1)$$ 
The collection $\mathcal{O}$ is closed under finite intersections, so it is a base of a topology on $X$. Such a topology is called the weak topology. The idea which stands behind this definition is simple: we want to obtain a topology which makes all the functions $f_j$ continuous. So the most natural way is to demand that $f^{-1}_j(O_j)$ are open sets and construct the topology via basis. There is also a very nice characterization of weak topology if we assume $X$ to be Hausdorff. 
Taking $X$ to be a normed space and $f_j\in X^*$ we have a theorem which gives a view on basis of neighbourhoods of $x_0$ for the weak topology. The sets given by
$$W^{x_{0}}_{f_1,…,f_k,\varepsilon}=\bigcap_{j=1}^k \{x\in X\,:\,|f_{j}(x)-f_j(x_{0})|<\varepsilon\}\qquad (2)$$ 
generate the basis of neighbourhoods. So the weak topology in this particular case may by introduced via (2). My intention is to define weak* topology via (3), i.e.,
$$W^{f_{0}}_{x_1,…,x_k,\varepsilon}=\bigcap_{j=1}^k \{f\in X^*\,:\,|f(x_j)-f_0 (x_{j})|<\varepsilon\}.\qquad (3)$$
In order to do so, I have to prove that for every weakly* open set $O$ containing $f_{0}$ there exists a set $W$ (at least one) of the form given by (3) such that $W\subset O$. Then I obtain the base of neighbourhoods and as a result the base of topology. But I have to take $O$ being a weak* neighbourhood. So what are the weak* open sets?
 A: For a normed space $X$ and $\mathcal{F}\subseteq X^*$ let $\sigma(X,\mathcal{F})$ be the smallest topology on $X$ such that all $f\in\mathcal{F}$ become continous (i.e. the intersection over all such topologies).Then a subbasis is given as in (1) and as in (2) of your question  a neighbourhoodbasis of $x_0\in X$ is given by the sets
$$W^{x_{0}}_{f_1,…,f_k,\varepsilon}=\bigcap_{j=1}^k \{x\in X\,:\,|f_{j}(x)-f_j(x_{0})|<\varepsilon\} \qquad (2)$$ where $k\in \mathbb{N}_0$, $f_j\in \mathcal{F}$ and $\varepsilon>0$.
(Same proof as for $\mathcal{F}=X^*$.)
There is the canonical linear isomorphism $Q: X\to Q(X)\subseteq X^{**},$ $ Q(x)(x^*)=x^*(x)$, so $Q(X)$ is the family of point evaluations. Now the weak* topology on $X^*$ is just the topology $\sigma (X^*,Q(X))$, which one often writes as $\sigma (X^*,X)$ because  $X\cong Q(X)$. By (2) a neighbourhoodbasis for $f_0\in X^*$ is given by the sets
$$W^{f_{0}}_{w_1,…,w_k,\varepsilon}=\bigcap_{j=1}^k \{f\in X^*\,:\,|w_{j}(f)-w_j(f_{0})|<\varepsilon\}$$
where $k\in \mathbb{N}_0$, $w_j\in Q(X)$ and $\varepsilon>0$. But since $Q$ is a bijection from $X$ to $Q(X)$, this collection of sets is the same as the collection of  sets
$$W^{f_{0}}_{x_1,…,x_k,\varepsilon}=\bigcap_{j=1}^k \{f\in X^*\,:\,|Q(x_{j})(f)-Q(x_j)(f_{0})|<\varepsilon\}\\
=\bigcap_{j=1}^k \{f\in X^*\,:\,|f(x_j)-f_{0}(x_j)|<\varepsilon\}$$
where $k\in \mathbb{N}_0$, $x_j\in X$ and $\varepsilon>0$, which is  exactly (3) in your question.
I hope this brings some clarification! Have fun!
