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infinite Galois theory
Let $L/F$ be a Galois extension, not necessarily finite dimensional.
1 Topology on the Galois group
Recall that the Galois group $G:=\operatorname{Gal}(L/F)$ of $L/F$ is the group of all field automorphisms $\sigma:L\longrightarrow L$ that restrict to the identity map on $F$, under the group operation of composition. In the case where the extension $L/F$ is infinite dimensional, the group $G$ comes equipped with a natural topology, which plays a key role in the statement of the Galois correspondence.
We define a subset $U$ of $G$ to be open if, for each $\sigma\in U$, there exists an intermediate field $K\subset L$ such that

If $\sigma^{{\prime}}$ is another element of $G$, and the restrictions $\sigma_{K}$ and $\sigma^{{\prime}}_{K}$ are equal, then $\sigma^{{\prime}}\in U$.
The resulting collection of open sets forms a topology on $G$, called the Krull topology, and $G$ is a topological group under the Krull topology. Another way to define the topology is to state that the subgroups $\operatorname{Gal}(L/K)$ for finite extensions $K/F$ form a neighborhood basis for $\operatorname{Gal}(L/F)$ at the identity.
2 Inverse limit structure
In this section we exhibit the group $G$ as a projective limit of an inverse system of finite groups. This construction shows that the Galois group $G$ is actually a profinite group.
Let $\mathcal{A}$ denote the set of finite normal extensions $K$ of $F$ which are contained in $L$. The set $\mathcal{A}$ is a partially ordered set under the inclusion relation. Form the inverse limit
$\Gamma:=\,\underset{\longleftarrow}{\lim}\,\operatorname{Gal}(K/F)\subset\prod% _{{K\in\mathcal{A}}}\operatorname{Gal}(K/F)$ 
consisting, as usual, of the set of all $(\sigma_{K})\in\prod_{K}\operatorname{Gal}(K/F)$ such that $\sigma_{{K^{{\prime}}}}_{K}=\sigma_{K}$ for all $K,K^{{\prime}}\in\mathcal{A}$ with $K\subset K^{{\prime}}$. We make $\Gamma$ into a topological space by putting the discrete topology on each finite set $\operatorname{Gal}(K/F)$ and giving $\Gamma$ the subspace topology induced by the product topology on $\prod_{K}\operatorname{Gal}(K/F)$. The group $\Gamma$ is a closed subset of the compact group $\prod_{K}\operatorname{Gal}(K/F)$, and is therefore compact.
Let
$\phi:G\longrightarrow\prod_{{K\in\mathcal{A}}}\operatorname{Gal}(K/F)$ 
be the group homomorphism which sends an element $\sigma\in G$ to the element $(\sigma_{K})$ of $\prod_{K}\operatorname{Gal}(K/F)$ whose $K$–th coordinate is the automorphism $\sigma_{K}\in\operatorname{Gal}(K/F)$. Then the function $\phi$ has image equal to $\Gamma$ and in fact is a homeomorphism between $G$ and $\Gamma$. Since $\Gamma$ is profinite, it follows that $G$ is profinite as well.
3 The Galois correspondence
Theorem 1 (Galois correspondence for infinite extensions).
Let $G$, $L$, $F$ be as before. For every closed subgroup $H$ of $G$, let $L^{H}$ denote the fixed field of $H$. The correspondence
$K\mapsto\operatorname{Gal}(L/K),$ 
defined for all intermediate field extensions $F\subset K\subset L$, is an inclusion reversing bijection between the set of all intermediate extensions $K$ and the set of all closed subgroups of $G$. Its inverse is the correspondence
$H\mapsto L^{H},$ 
defined for all closed subgroups $H$ of $G$. The extension $K/F$ is normal if and only if $\operatorname{Gal}(L/K)$ is a normal subgroup of $G$, and in this case the restriction map
$G\longrightarrow\operatorname{Gal}(K/F)$ 
has kernel $\operatorname{Gal}(L/K)$.
Theorem 2 (Galois correspondence for finite subextensions).
Let $G$, $L$, $F$ be as before.

Every open subgroup $H\subset G$ is closed and has finite index in $G$.

If $H\subset G$ is an open subgroup, then the field extension $L^{H}/F$ is finite.

For every intermediate field $K$ with $[K:F]$ finite, the Galois group $\operatorname{Gal}(L/K)$ is an open subgroup of $G$.
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References
Ca you please add some references?
An open invitation
Riemann Hypothesis:
My gut feeling with regard to RH is that it is true.I may be wrong but to prove it it may need an OR approach i.e. only a team consisting of a)analytical number theorists b)algebraists c)topologists d) geometers e) programmers and, perhaps a physicist
jointly attacking the problem can crack it.
Perhaps PM can take the initiative in forming such a team.
A.K.Devaraj