The product topology of weak topologies is the same as the weak topology of the product space? I am reading Brezis' Functional Analysis. On page 62, in Theorem 3.10, it says 

... $E \times F$ equipped with the product topolgy $\sigma(E, E^\star) \times \sigma(F, F^\star)$, which is clearly the same as $\sigma(E\times F, (E \times F)^\star)$. 

Here, $E$ and $F$ are Banach spaces and $\sigma(E, E^\star)$ and $\sigma(F, F^\star)$ are their weak topologies respectively. 
Actually, I showed that $\sigma(E, E^\star) \times \sigma(F, F^\star) \subset \sigma(E\times F, (E \times F)^\star)$ in the following way. 

Let $\pi: E \times F \to E$ with $\pi(x,y) = x$ and $f_1 \in E^\star$. Then, since $f_1 \circ \pi \in (E \times F)^\star$, $(f \circ \pi)^{-1}(O_1) = \pi^{-1}(f_1^{-1}(O_1)) = f_1^{-1}(O_1) \times F \in \sigma(E\times F, (E \times F)^\star)$ for an open set $O_1$ in $\mathbb{R}$. 

In the same manner, 

if $f_2 \in F^\star$, then $f_2^{-1}(O_2) \times F \in \sigma(E\times F, (E \times F)^\star)$ for an open set $O_2$ in $\mathbb{R}$. Therefore, $f_1^{-1}(O_1) \times f_2^{-1}(O_2) \in \sigma(E\times F, (E \times F)^\star)$. Since $f_1^{-1}(O_1) \times f_2^{-1}(O_2)$ is a basis member of $\sigma(E, E^\star) \times \sigma(F, F^\star)$, $\sigma(E, E^\star) \times \sigma(F, F^\star) \subset \sigma(E\times F, (E \times F)^\star)$.

However, I cannot show $\sigma(E, E^\star) \times \sigma(F, F^\star) \supset \sigma(E\times F, (E \times F)^\star)$. How can we show this? Any help would be really appreciated!
 A: We may prove this by first principles: Let $U \in \sigma(E\times F, (E\times F)^{\ast})$ be an open set in $E\times F$ and $(x_0,y_0) \in U$. Then, there exist finitely many bounded linear functionals $\varphi_i : E\times F \to \mathbb{C}, 1\leq i\leq k$ and finitely many positive real numbers $\epsilon_i > 0$ so that the set
$$
W := \bigcap_{i=1}^n \{(x,y) \in E\times F : |\varphi_i(x,y) - \varphi_i(x_0,y_0)| < \epsilon_i\}
$$
is contained in $U$. Let $\iota_E : E\to E\times F$ and $\iota_F : F\to E\times F$ denote the inclusions $x\mapsto (x,0)$ and $y\mapsto (0,y)$ respectively. For each $1\leq i\leq k$, define
$$
A_i := \{x \in E : |\varphi_i\circ \iota_E(x) - \varphi_i\circ \iota_E(x_0)| < \epsilon_i/2\}
$$
and
$$
B_i := \{y\in F : |\varphi_i\circ \iota_F(y) - \varphi_i\circ\iota_F(y_0)| < \epsilon_i/2\}
$$
Now $A_i \in \sigma(E,E^{\ast})$ and $B_i \in \sigma(F,F^{\ast})$, so if
$$
A := \bigcap_{i=1}^n A_i, \text{ and } B := \bigcap_{i=1}^n B_i
$$
Then
$$
A\times B \in \sigma(E,E^{\ast})\times \sigma(F,F^{\ast})
$$
and if $(x,y) \in A\times B$, then for any $1\leq i\leq k$,
$$
|\varphi_i(x,y) - \varphi_i(x_0,y_0)| < \epsilon_i
$$
Hence, $A\times B \subset W\subset U$. Thus, $U \in \sigma(E,E^{\ast})\times \sigma(F,F^{\ast})$ as required.
