I am trying to get my head around the weak topologies and i came across this problem:Let $X$ be a Banach space and $$ q:X\longrightarrow X^{**},q_{x}(x^{*})=x^{*}(x)$$.This is called the natural embeeding on the bidual space.
Now we need to prove that the weak topology on $X$ coincides with the topology induced by $(X^{**},w^{*}) $ on $X$ through $q$.
Here is my attempt:The topology induced by this mapping has a base formed by the sets $q^{-1}(V)$,$V$ is $w^{*}$ neighbourhood of $X^{**}$.$$V=\{x^{**}\in X^{**}: |x^{**}(x^{*}_i)|<\epsilon,i=1,..,n \}$$ for some $\epsilon>0 ,x^{*}_i\in X^{*}$ and $$q^{-1}(V)=\{x\in X: q_x\in V\}=\{x\in X: |x^{*}_i(x)|<\epsilon,i=1,...,n\}$$ with the last set being a weak neighbourhood of the original space. Is it as trivial as it seems or am i missing something serious here??? Any hints would be great.
You're not missing anything: it really is as trivial as it seems. The basic open neighborhoods are just exactly the same when you identify $X$ with $q(X)$.