Define the set$$\Delta :=\left \{x=(x_n)_{n\in \mathbb{N}}\in \mathbb{R}^\mathbb{N}:x_n\geq 0\,\,\forall n\in \mathbb{N},\sum \limits _{n=1}^\infty x_n\leq 1\right \}.$$Prove that the product topology on the infinite-dimensional simplex agrees with the weak$^*$ topology it inherits as a subset of $\ell ^1=c_0^*$.
Here is what I have so far: Let $\tau _p$ be the product topology on $\Delta$ and $\tau ^*$ be the weak$^*$ topology on $\Delta$. For every linear functional $y^*\in c_0^*$, there exists a unique element $y\in \ell ^1$ such that$$y^*(x)=\sum \limits _{n=1}^\infty x_ny_n$$for all $x\in c_0$. We know that in general there exists an isometric embedding $\iota :X\to X^{**}$ such that for each $x\in X$ we have$$\iota (x)(x^*)=x^*(x)$$for all $x^*\in X^*$. In particular, for $X=c_0$ we have$$\iota (x)(y^*)=y^*(x)=\sum \limits _{n=1}^\infty x_ny_n$$for all $y^*\in c_0^*$. Now, if $\pi _k:\ell ^1\to \mathbb{R}$ is the projection over the $k$-th coordinate and $e^{(k)}:=(\delta _{k,n})_{n\in \mathbb{N}}$, where $\delta _{k,n}$ is the Kronecker $\delta$ (i.e. $e^{(k)}$ is the vector in $\ell ^1$ that has a $1$ in the $k$-th position and $0$ elsewhere), we have$$\iota \left (e^{(k)}\right )(y^*)=\sum \limits _{n=1}^\infty \delta _{k,n}y_n=y_k=\pi _k(y)$$for all $y^*\in c_0^*$, hence $\color{red}{(1)}$ we have $\iota \left (e^{(k)}\right )=\pi _k$. Therefore, $\tau _p\subset \tau ^*$.
I'm not sure if $\color{red}{(1)}$ is written correctly. What is the right way of writing/proving that $\iota \left (e^{(k)}\right )=\pi _k$? Also, I don't know how to prove that $\tau ^*\subset \tau _p$ (clearly, we have to prove that every element of $\iota (c_0)$ is continuous in $\tau _p$, but I haven't figured that out yet). In case it is useful, I know that the product topology on $\mathbb{R}^\mathbb{N}$ is induced by the distance$$d(x,y)=\sum \limits _{n=1}^\infty \frac{2^{-n}|x_n-y_n|}{1+|x_n-y_n|}$$ and that $\Delta$ is compact in $\mathbb{R}^\mathbb{N}$.
Another possible idea is to observe that the product topology is also induced by the distance$$D(x,y)=\sup \limits _{n\in \mathbb{N}}\left \{\frac{\min \{|x_n-y_n|,1\}}{n}\right \},$$hence we only need to prove that given $x\in c_0$, we have that for all $\varepsilon >0$ there exists $\delta >0$ such that$$D(y,0)<\delta \implies \left |\sum \limits _{n=1}^\infty x_ny_n\right |<\varepsilon$$i.e.$$\frac{y_n}{n}<\delta \quad \forall n\in \mathbb{N}\implies \left |\sum \limits _{n=1}^\infty x_ny_n\right |<\varepsilon .$$