Let $X$ be a Banach space over $\mathbf{R}$ and denote $U = X'$, the dual of $X$. As a part of Banach-Alaoglu i want to show that the embedding of the set $B = \{u\in U\mid \|u\|\leq 1\}$ into the set
$$P = \prod_{x\in X}[-|x|,|x|]$$
given by the map $u\mapsto (u(x))_{x\in X}$ preserves the topological structure. Here $P$ is given the product topology. In the words of the author
...the weak* topology of $B$ is the same as the inherited topology from $P$ under the embedding.
My Question: Am i correct in that I want to show that if $U$ is an open subset of $B$ if and only if $\{(u(x))_{x\in X}\mid u\in U\}$ is open in $P$?
Since my claim was verified by @Berci I proceed with showing that the statement is valid: $U$ is an open subset of $B$ in the weak* topology if and only $Y = \{(u(x))_{x\in X}\mid u\in U\}$ is open in $P$.
A subbasis element of $B$ is given by
$$S = \{u\in U\mid |u(x)-u_0(x)|<\varepsilon\}$$
however that is the set $\pi_x^{-1}(u_0(x)-\varepsilon,u_0(x)+\varepsilon)\cap Y$ and therefore it is open in $Y$. But then finite intersections followed by arbitrary unions of elements of the type $S$ is open in $Y$ and hence if a set is open in $B$ then it is open in $Y$.
Conversely a subbasis element of $\{(u(x))_{x\in X}\mid u\in U\}$ is given by $S' = Y\cap \pi_x^{-1}((a,b))$. Say $u_0$ belongs to the intersection $S'$ so that $u_0(x)\in (a,b)$ then finding $\varepsilon>0$ so that $(u_0(x)-\varepsilon,u_0(x)+\varepsilon)$ we see that
$$\{u\in U\mid |u(x)-u_0(x)|<\varepsilon\}\subseteq S'.$$
This shows that $S'$ is open in the weak* topology on $B$.