I am reading the proof of
Alaoglu's theorem: Let $X$ be a normed linear space. Then the closed unit ball of $X^*$ is weak-* compact.
Part of the proof:
Proof: Let $B_{X^*}$ denote the closed unit ball of $X^*$. Then if $f\in B_{X^*}$ we have that $\|f\|\le 1$ and so for every $x\in X$, $|f(x)|\le \|f\| \|x\| \le \|x\|$. So for every $x\in X$ and every $f\in B_{X^*}$ we ave that $f(x) \in [-\|x\|, \|x\|]$. For each $x\in X$ let $I_x = [-\|x\|, \|x\|]$ and let $$\tag{2} P = \Pi_{x\in X} I_x.$$ Then $$\tag{3} B_{X^*} \subset P.$$
In (3), we claim that $B_{X^*} \subset P$. I was a bit confused with the subset notation.
$B_{X^*}$ is a set of bounded linear functionals with norm $\leq 1$. $P$ is a Cartesian product of closed interval. I am wondering what is the correct way to understand the equation?
By definition, $$\prod_xI_x=\Big\{f:X\to\bigcup_xI_x:\ f(x)\in I_x\ \text{ for all }x\Big\}.$$ So indeed $B_{X^*}\subset P$.