Consider $V$ a normed vector space, and equip its topological dual $V^*$ with the usual operator norm. Consider also the weak* topology on $V^*$ defined by the seminorms $(p_A)_{A\subset V \text{finite}}$ defined as follows : $\forall f \in V \quad p_A(f) = \sup_{x\in A} |f(x)|$
It says in my lecture notes that $\bar B(0,1)$ in $V^*$ is compact for the weak topology. And the proof uses Tychonoff's theorem. What I do not understand, is that the justification for using the theorem is that $\bar B(0,1)$ can be injected into $[0,1]^{[0,1]}$. I really don't see how this can canonically be done.
$\bar{B}(0,1)$ is weak-$*$ compact. This result is called Banach-Alaoglu theorem.
First, try to understand that the weak-$*$ topology of $V^*$ is precisely the product topology of $R^V$ induced on the subset $V^*$.
Second, let us consider $K:=\underset{x\in V}\prod [-\Vert x\Vert,\Vert x\Vert]$. Due to the Tychonoff's theorem, since $[-\Vert x\Vert,\Vert x\Vert]$ is compact in $R$, it turns out that $K$ is compact with respect to the product topology of $R^V$.
Finally, try to prove that $\bar{B}(0,1)$ is a closed subset of $K$. Consequently, $\bar{B}(0,1)$ will be compact for the product topology induced on $V^*$, that is, the weak-$*$ topology.
Edited: I have changed $[0,1]^V$ by $K:=\underset{x\in V}\prod [-\Vert x\Vert,\Vert x\Vert]$.