[proofplan]
We prove the Fundamental Theorem of Infinite Galois Theory by reducing to the finite case via the inverse limit structure of $\operatorname{Gal}(M/K)$. The key step is showing that the map $L \mapsto \operatorname{Gal}(M/L)$ sends subextensions to closed subgroups and has inverse $H \mapsto M^H$. Injectivity follows from the finite Galois theory applied to finite subextensions. Surjectivity (every closed subgroup arises as $\operatorname{Gal}(M/L)$) uses the fact that a closed subgroup $H$ equals the intersection of its open supergroups, each of which corresponds to a finite subextension by the finite theory. The additional characterisations of finite and Galois subextensions follow from standard group-theoretic arguments.
[/proofplan]
[step:Show $\operatorname{Gal}(M/L)$ is a closed subgroup for every subextension $L$]
Let $L/K$ be a subextension of $M/K$. Write $L = \bigcup_{\alpha \in L} K(\alpha)$ and for each $\alpha \in L$, let $L_\alpha$ be a finite Galois extension of $K$ contained in $M$ that contains $\alpha$ (such an extension exists because $M/K$ is Galois). Then:
\begin{align*}
\operatorname{Gal}(M/L) = \bigcap_{\alpha \in L} \operatorname{Gal}(M/L_\alpha) \cap \{\sigma \in \operatorname{Gal}(M/K) : \sigma(\alpha) = \alpha\}.
\end{align*}
More directly: $\operatorname{Gal}(M/L) = \{\sigma \in \operatorname{Gal}(M/K) : \sigma|_L = \operatorname{id}_L\}$. For each $\alpha \in L$, the set $\{\sigma : \sigma(\alpha) = \alpha\}$ is the preimage of $\{\operatorname{id}|_{L_\alpha}\}$ under the continuous restriction map $\operatorname{Gal}(M/K) \to \operatorname{Gal}(L_\alpha/K)$, hence closed (as the preimage of a point in a discrete space). Therefore:
\begin{align*}
\operatorname{Gal}(M/L) = \bigcap_{\alpha \in L} \{\sigma : \sigma(\alpha) = \alpha\}
\end{align*}
is an intersection of closed sets, hence closed. It is a subgroup because the identity fixes $L$, and if $\sigma, \tau$ fix $L$ then so do $\sigma\tau$ and $\sigma^{-1}$.
[/step]
[step:Show $M^{\operatorname{Gal}(M/L)} = L$ for every subextension $L$]
The inclusion $L \subseteq M^{\operatorname{Gal}(M/L)}$ is immediate: every element of $L$ is fixed by every $\sigma \in \operatorname{Gal}(M/L)$.
For the reverse inclusion, let $\beta \in M \setminus L$. We must find $\sigma \in \operatorname{Gal}(M/L)$ with $\sigma(\beta) \neq \beta$. Let $N$ be a finite Galois extension of $K$ contained in $M$ that contains $\beta$ and a finite generating set of $L \cap N$ over $K$. By the finite Galois correspondence applied to $N/K$, we have $N^{\operatorname{Gal}(N/L \cap N)} = L \cap N$. Since $\beta \notin L$, we have $\beta \notin L \cap N$ (as $\beta \in N$, so $\beta \in L$ iff $\beta \in L \cap N$). Therefore there exists $\tau \in \operatorname{Gal}(N/L \cap N)$ with $\tau(\beta) \neq \beta$. Extend $\tau$ to an automorphism $\sigma \in \operatorname{Gal}(M/K)$ (possible since $M/K$ is Galois and $N/K$ is a finite Galois subextension, so the restriction map $\operatorname{Gal}(M/K) \to \operatorname{Gal}(N/K)$ is surjective). Since $\tau$ fixes $L \cap N$ and $\sigma|_N = \tau$, we need $\sigma$ to fix all of $L$.
We refine: choose $N$ to be a finite Galois extension of $K$ in $M$ containing both $\beta$ and enough elements of $L$ so that $L \cap N$ generates the relevant constraints. More precisely, for any finite Galois $N/K$ in $M$, the restriction map gives a surjection $\operatorname{Gal}(M/K) \twoheadrightarrow \operatorname{Gal}(N/K)$ with kernel $\operatorname{Gal}(M/N)$. The image of $\operatorname{Gal}(M/L)$ under this restriction is $\operatorname{Gal}(N / L \cap N)$ (by the finite theory, since $\operatorname{Gal}(M/L)|_N = \operatorname{Gal}(N / L \cap N)$). So there exists $\sigma \in \operatorname{Gal}(M/L)$ with $\sigma|_N = \tau$, giving $\sigma(\beta) = \tau(\beta) \neq \beta$.
Hence $M^{\operatorname{Gal}(M/L)} = L$.
[guided]
The surjectivity of the restriction $\operatorname{Gal}(M/L) \to \operatorname{Gal}(N / L \cap N)$ is a key point. Why does it hold? Consider the diagram:
\begin{align*}
\operatorname{Gal}(M/K) &\twoheadrightarrow \operatorname{Gal}(N/K) \\
\cup & \qquad \cup \\
\operatorname{Gal}(M/L) &\to \operatorname{Gal}(N / L \cap N)
\end{align*}
The top arrow is surjective (since $M/K$ is Galois and $N \subseteq M$). The image of $\operatorname{Gal}(M/L)$ under restriction to $N$ consists of all $\tau \in \operatorname{Gal}(N/K)$ that fix $L \cap N$, which is exactly $\operatorname{Gal}(N / L \cap N)$. The surjectivity follows because for any $\tau \in \operatorname{Gal}(N / L \cap N)$, lifting $\tau$ to $\sigma \in \operatorname{Gal}(M/K)$ and noting that $\sigma$ fixes $L \cap N$ — but we need $\sigma$ to fix all of $L$. This requires a more careful argument using the profinite structure.
In fact, the statement follows from a compactness argument: the set $\operatorname{Gal}(M/L) \cap \operatorname{res}_N^{-1}(\{\tau\})$ is an intersection of the closed set $\operatorname{Gal}(M/L)$ with the open set (hence also closed, by finite index) $\operatorname{res}_N^{-1}(\{\tau\})$. As $\tau$ ranges over $\operatorname{Gal}(N / L \cap N)$, these sets partition $\operatorname{Gal}(M/L)$. Since $\operatorname{Gal}(M/L)$ is nonempty (it contains $\operatorname{id}$ at minimum), and the partition into cosets of $\operatorname{Gal}(M/NL)$ gives a bijection between $\operatorname{Gal}(M/L)/\operatorname{Gal}(M/NL) \cong \operatorname{Gal}(NL/L) \cong \operatorname{Gal}(N / N \cap L)$, the restriction is surjective.
[/guided]
[/step]
[step:Show every closed subgroup $H$ has $\operatorname{Gal}(M/M^H) = H$]
Let $H$ be a closed subgroup of $\operatorname{Gal}(M/K)$ and set $L = M^H$. The inclusion $H \subseteq \operatorname{Gal}(M/L)$ is immediate: every $\sigma \in H$ fixes $M^H = L$.
For the reverse inclusion $\operatorname{Gal}(M/L) \subseteq H$, let $\sigma \in \operatorname{Gal}(M/K) \setminus H$. Since $H$ is closed and $\operatorname{Gal}(M/K)$ is Hausdorff, $\{\sigma\}$ and $H$ can be separated by open sets. More concretely, since $\sigma \notin H$ and $H$ is closed, there is a basic open neighbourhood $\sigma \cdot \operatorname{Gal}(M/N)$ of $\sigma$ (for some finite Galois $N/K$) that is disjoint from $H$. This means $\sigma|_N \notin H|_N$ (the image of $H$ under restriction to $N$).
Now $H|_N$ is a subgroup of $\operatorname{Gal}(N/K)$. By the finite Galois correspondence, $N^{H|_N}$ is the fixed field of $H|_N$ in $N$, and $H|_N = \operatorname{Gal}(N / N^{H|_N})$. Since $H$ fixes $L = M^H$, we have $H|_N \subseteq \operatorname{Gal}(N / L \cap N)$, so $L \cap N \subseteq N^{H|_N}$.
Since $\sigma|_N \notin H|_N = \operatorname{Gal}(N / N^{H|_N})$, there exists $\alpha \in N^{H|_N}$ with $\sigma(\alpha) \neq \alpha$. But $L \cap N \subseteq N^{H|_N}$, so $\alpha \in N^{H|_N}$, and since $H$ fixes $\alpha$ (all elements of $H$ restrict to $H|_N$ on $N$, which fixes $N^{H|_N}$), we have $\alpha \in M^H = L$. Therefore $\sigma$ does not fix $L$, so $\sigma \notin \operatorname{Gal}(M/L)$.
This shows $\operatorname{Gal}(M/K) \setminus H \subseteq \operatorname{Gal}(M/K) \setminus \operatorname{Gal}(M/L)$, i.e., $\operatorname{Gal}(M/L) \subseteq H$, giving $\operatorname{Gal}(M/L) = H$.
[guided]
The argument uses the closure of $H$ in an essential way. If $H$ were merely a subgroup (not closed), we could only conclude $H \subseteq \operatorname{Gal}(M/M^H)$, but not equality. The closure hypothesis allows us to separate $\sigma \notin H$ from $H$ using an open set, which translates into a finite Galois extension where $\sigma$ and $H$ differ.
This is where infinite Galois theory diverges from the finite case: not every subgroup of $\operatorname{Gal}(M/K)$ corresponds to a subextension — only the closed subgroups do. A non-closed subgroup $H$ gives $\operatorname{Gal}(M/M^H) = \overline{H}$ (the closure of $H$), which is strictly larger than $H$.
[/guided]
[/step]
[step:Establish the bijection and prove the additional properties]
**Bijection:** The maps $L \mapsto \operatorname{Gal}(M/L)$ and $H \mapsto M^H$ are mutually inverse bijections between subextensions of $M/K$ and closed subgroups of $\operatorname{Gal}(M/K)$:
- $M^{\operatorname{Gal}(M/L)} = L$ (shown above).
- $\operatorname{Gal}(M/M^H) = H$ for closed $H$ (shown above).
**Finite subextensions correspond to open subgroups:** $L/K$ is finite if and only if $\operatorname{Gal}(M/L)$ has finite index in $\operatorname{Gal}(M/K)$, which (in a compact group) is equivalent to $\operatorname{Gal}(M/L)$ being open. The index satisfies $(\operatorname{Gal}(M/K) : \operatorname{Gal}(M/L)) = [L:K]$ when $L/K$ is Galois, and more generally $(\operatorname{Gal}(M/K) : \operatorname{Gal}(M/L)) = [L:K]$ since the cosets of $\operatorname{Gal}(M/L)$ in $\operatorname{Gal}(M/K)$ biject with the $K$-embeddings of $L$ into $M$, of which there are $[L:K]$ (by separability). In a compact topological group, a subgroup of finite index is open if and only if it is closed, and $\operatorname{Gal}(M/L)$ is always closed.
**Galois subextensions correspond to normal subgroups:** $L/K$ is Galois if and only if $\operatorname{Gal}(M/L)$ is normal in $\operatorname{Gal}(M/K)$. This follows from the same argument as in finite Galois theory: $\sigma \operatorname{Gal}(M/L) \sigma^{-1} = \operatorname{Gal}(M/\sigma(L))$, so $\operatorname{Gal}(M/L)$ is normal iff $\sigma(L) = L$ for all $\sigma \in \operatorname{Gal}(M/K)$, iff $L/K$ is normal (i.e., Galois, since $L/K$ is automatically separable as a subextension of a Galois extension).
When $L/K$ is Galois, the restriction map $\operatorname{Gal}(M/K) \to \operatorname{Gal}(L/K)$, $\sigma \mapsto \sigma|_L$, is a continuous surjective group homomorphism with kernel $\operatorname{Gal}(M/L)$. By the first isomorphism theorem for topological groups, $\operatorname{Gal}(M/K)/\operatorname{Gal}(M/L) \cong \operatorname{Gal}(L/K)$ as topological groups (the quotient topology on the LHS coincides with the Krull topology on the RHS, since both are compact Hausdorff and the bijection is continuous, hence a homeomorphism by compactness).
[/step]
